SHIVAJI UNIVERSITY, KOLHAPUR
CENTRE FOR DISTANCE EDUCATION
Complex Analysis
(Mathematics)
For
M. Sc.-I
Copyright ©
Registrar,
Shivaji University,
Kolhapur. (Maharashtra)
First Edition 2009
Prescribed for M. Sc. Part-I
All rights reserved, No part of this work may be reproduced in any form by mimeography or
any other means without permission in writing from the Shivaji University, Kolhapur (MS)
Copies : 500
Published by:
Dr. D. T. Shirke
I/c. Registrar,
Shivaji University,
Kolhapur-416 004
Printed by :
Shri. A. S. Mane,
I/c. Superintendent,
Shivaji University Press,
Kolhapur-416 004
Cover Design by :
Pratik Printing Services, Kolhapur.
ISBN-978-81-8486-136-5
H
Further information about the Centre for Distance Education & Shivaji University may be
obtained from the University Office at Vidyanagar, Kolhapur-416 004, India.
H
This material has been produced with the developmental grant from DEC-IGNOU, New
Delhi.
(ii)
Centre for Distance Education
Shivaji University, Kolhapur
n EXPERT COMMITTEE n
Dr. D. T. Shirke
Prof. (Dr.) A. A. Dange
Ag. Vice-Chancellor,
Shivaji University, Kolhapur
I/c. Registrar,
Shivaji University, Kolhapur.
n ADVISORY COMMITTEE n
Prof. (Dr.) A. A. Dange
Ag. Vice-Chancellor,
Shivaji University, Kolhapur.
Dr. D. T. Shirke
I/c. Registrar,
Shivaji University, Kolhapur.
Dr. B. M. Hirdekar
Controller of Examination
Shivaji University, Kolhapur.
Shri. B. S. Patil
Finance and Accounts Officer,
Shivaji University, Kolhapur.
Dr. (Smt.) Vasanti Rasam,
Dean, Faculty of Social Sciences,
Shivaji University, Kolhapur.
Prof. (Dr.) U. B. Bhoite,
Lal Bahadur Shastri Marg,
Bharati Vidyapeeth, Pune.
Prof. (Dr.) V. S. Patil,
Dean, Faculty of Commerce,
Shivaji University, Kolhapur.
Prof. (Dr.) A. N. Joshi,
Director, School of Education,
Y. C. M. O. U. Nashik.
Dr. T. B. Jagtap,
Dean, Faculty of Science,
Shivaji University, Kolhapur.
Shri. J. R. Jadhav,
Dean, Faculty of Arts & Fine Arts,
Shivaji University, Kolhapur.
Dr. K. N. Sangale,
Dean, Faculty of Education,
Shivaji University, Kolhapur.
Prof. Dr. (Smt.) Cima Yeole
(Member Secretary)
Director, Centre for Distance Education,
Shivaji University, Kolhapur.
Prof. (Dr.) S. A. Bari
Director, Distance Education,
Kuvempu University, Karnataka.
n B. O. S. MEMBERS OF MATHEMATICS n
Chairman- Prof. S. R. Bhosale
P.D.V.P. Mahavidyalaya, Tasgaon, Dist. Sangli.
l
l
l
l
Dr. M. S. Choudhari
Head, Dept. of Mathematics,
Shivaji University, Kolhapur.
Dr. H. T. Dinde
Karmveer Bhaurao Patil College, Urun-Islampur,
Tal. Walwa, Dist. Sangli.
l
l
l
Dr. T. B. Jagtap
Yashwantrao Chavan Institute of Science, Satara.
Shri. L. B. Jamale
Krishna Mahavidyalaya, Rethare Bk., Karad,
Dist. Satara.
l
Dr. A. D. Lokhande
Yashavantrao Chavan Warana Mahavidyalaya,
Warananagar.
Prof. S. P. Patankar
Vivekanand College, Kolhapur.
Prof. V. P. Rathod
Dept. of Mathematics, Gulbarga University,
Gulbarga, (Karnataka State.)
Prof. S. S. Benchalli
Dept. of Mathematics,Karnataka University, Dharwad.
l
(iii)
Shri. Santosh Pawar,
1012, 'A' Ward Sadashiv Jadhav, Housing Society,
Radhanagari Road, Kolhapur-416 012.
Centre for Distance Education
Shivaji University,
Kolhapur.
Complex Analysis
Writing Team
Dr. S. R. Chaudhari
Dept. of Mathematics,
Shivaji University, Kolhapur.
(Maharashtra)
Dr. U. H. Naik
Department of Mathematics,
Shivaji University, Kolhapur.
(Maharashtra)
n
Editor n
Dr. S. R. Chaudhari
Dr. U. H. Naik
Department of Mathematics,
Shivaji University, Kolhapur.
(Maharashtra)
Department of Mathematics,
Shivaji University, Kolhapur.
(Maharashtra)
(iv)
Preface
The Shivaji University, Kolhapur has established the Distance Education Centre
for external students from the year 2007-08, with the goal that, those students who are
not able to complete their studies regularly, due to unavoidable circumstances, they
must be involved in the main stream by appearing externally. The centre is trying hard
to provide notes to those aspirants by entrusting the task to experts in the subjects to
prepare the Self Instructional Material (SIM). Today we are extremely happy to present
a book on Complex Analysis for M. Sc. Mathematics students as SIM prepared by us.
The SIM is prepared strictly according to syllabus and we hope that the exposition of
the material in the book will meet the needs of all students.
This book introduces the students the most interesting and beautiful analysis viz.
Complex Analysis. As a matter of fact Complex Analysis is a hard analysis, but it is
truly a beautiful Analysis. The first topic is an introduction to Complex analysis. The
second unit deals with Mobius transformations. The third unit introduces the reader to
the notion of complex integration. Fundamental theorem of algebra and maximum
modulus theorem are the results covered in the unit four. Unit five and six cover
concept of winding number, Cauchy's integral theorem, Open mapping theorem and
Goursat theorem. Laurent series development, Residue theorem with its application
to evaluation of Real integrals, Rouche's theorem and Maximum Modulus theorem are
the results contained in last two units.
We owe a deep sense of gratitude to the Ag. Vice-Chancellor Dr. A. A. Dange who
has given impetus to go ahead with ambitious projects like the present one. Dr. S. R.
Chaudhari and Dr. U. H. Naik have to be profusely thanked for the ovation for they have
poured to prepare the SIM on Complex Analysis (M.Sc. Mathematics).
We also thank Professor M. S. Chaudhary, Head of the Department of Mathematics,
Shivaji University, Kolhapur, Director of Distance Education Mode Dr. Mrs. Cima Yeole
and Deputy Director, Shri. Sanjay Ratnaparakhi for their help and keen interest in
completion of the SIM. Thanks are also due to Mr. Girish Shelke who had taken pains
in typing the manuscript and Mr. Sachin Kadam for providing printing copy of the
manuscript neatly and correctly.
Prof. S. R. Bhosale
Chairman BOS in Mathematics
Shivaji University, Kolhapur-416004.
(v)
M. Sc. (Mathematics)
Complex Analysis
Contents
Unit-1 :
Complex Numbers
1
Unit-2 :
Mobius Transformations
16
Unit-3 :
Complex Integration
25
Unit-4 :
Fundamental Theorem of Algebra and
Maximum Modulus Theorem
42
Unit-5 :
Winding Numbers and Cauchys Integral Theorem
54
Unit-6 :
Open Mapping Theorem and Goursat Theorem
71
Unit-7 :
Laurent Series Development and Residue Theorem
78
Unit-8 :
Rouche's Theorem and Maximum Modulus Theorem
97
(vii)
UNIT - I
COMPLEX NUMBERS
Introduction
We know that in the real number system
, the equation x 2 + a = 0 has no solution.
This leads to introduction of complex number system in which equations of the form
x 2 + a = 0 , where a > 0 , have solutions. This chapter introduces complex numbers, their
representation and basic properties.
Definition 1 The complex numbers can be defined as pair of real numbers
= {( x, y ) : x, y ∈ } .
Equipped
with
addition
( x , y ) + ( a, b ) = ( x + a , y + b )
and
multiplication being defined as ( x, y )(a, b) = ( xa − yb, xb + ya ) .
One reason to believe that the definitions of these binary operations are “good” is that
is an extension of
, in the sense that the complex numbers of the form ( x, 0) behave just
like real numbers; that is, ( x, 0) + (a, 0) = ( x + a, 0) and ( x, 0)(a, 0) = ( xa, 0) . So we can think
of the real numbers being embedded in
as those complex numbers whose second
coordinate is zero. The following basic results states the algebraic structure that we
established with our definitions. Its proof is straightforward but nevertheless a good exercise.
1. Commutative law for addition : z1 + z2 = z2 + z1 .
2. Associative law for addition : z1 + ( z2 + z3 ) = ( z1 + z2 ) + z3 .
3. Additive identity : There is a complex number z ' such that
z + z0 = z for all
complex number z . The number z0 is an ordered pair ( 0, 0 ) .
4. Additive inverse : For any complex number z there is a complex number − z such
that z + ( − z ) = ( 0, 0 ) . The number − z is
( − x, − y ) .
5. Commutative law for multiplication : z1 z2 = z2 z1 .
6. Associative law for multiplication : z1 ( z2 z3 ) = ( z1 z2 ) z3 .
1
7. Multiplicative identity : There is a complex number z ' such that
zz ' = z for all
complex number z . The number z ' is an ordered pair (1, 0 ) .
8. Multiplicative inverse : For any non-zero complex number z there is a complex
 x
−y 
, 2
.
number z −1 such that zz −1 = (1, 0 ) . The number z −1 is  2
2
2 
x +y x +y 
9. The distributive law : z1 ( z2 + z3 ) = z1 z2 + z1 z3 .
If we write x for the complex number
isomorphism of
into
( x, 0 ) .
so we may consider
This mapping x → ( x, 0 ) defines a field
as a subset of
.
If we put i = ( 0,1) , then z = ( x, y ) = ( x, 0 ) + ( 0, y ) = ( x, 0 ) + ( 0,1)( y, 0 ) = x + iy .
Let z = x + iy , x, y ∈
, then x and y are called the real and imaginary parts of z and denote
this by x = Re z , y = Im z . If x = 0 , the complex number z is called purely imaginary and if
y = 0 , then z is real. Note that zero is the only number which is at once real and purely
imaginary. Two complex numbers are equal iff they have the same real part and the same
imaginary part.
Complex Plane or Argand plane : The number z = ( x, y ) = x + iy can be identified with the
unique point
( x, y )
in the plane
2
. The plane
2
representing the complex numbers is
called the complex plane. The x-axis is also called the real axis and the y-axis is called the
imaginary axis.
Definition 2 Let z = x + iy , x, y ∈
then the complex number x − iy is called the conjugate
of z and is denoted by z .
2
Following are the basic properties of conjugates.
1. Re z =
z+z
z−z
and Im z =
.
2
2i
2. z is real iff z = z .
3. z1 + z2 = z1 + z2
4. z1 z2 = z1 z2 .
z  z
5.  1  = 1 if z2 ≠ 0 .
 z2  z 2
6. z = z .
Definition 3 Let z = x + iy , x, y ∈
then modulus or absolute value of z is a non-negative
1
real number denoted by z and is given by z = ( x 2 + y 2 ) 2 . The number z is the distance
between the origin and the point ( x, y ) .
Following are the basic properties of Modulus.
2
1. z = z z
2. z1 z2 = z1 z2
3.
z
z1
= 1 if z2 ≠ 0 .
z2
z2
4. z1 z2 = z1 z2 .
5. z = z .
6. x = Re ( z ) ≤ z and y = Im ( z ) ≤ z .
7. z1 + z2 ≤ z1 + z2 .
8. z1 − z2 ≥ z1 − z2 .
9. Let z1 = x1 + iy1 , z2 = x2 + iy2 then
z1 − z2 = ( x1 − x2 ) + i ( y1 − y2 )
1
2
2
= ( x1 − x2 ) + ( y1 − y2 )  2 which is the distance between the


points ( x1 , y1 ) , ( x2 , y2 ) . Hence distance between the points z1 and z2 is given by z1 − z2 .
3
Polar representation of complex numbers
Consider the point
z = x + iy in the complex plane
. This point has polar co-
ordinates ( r ,θ ) where x = r cos θ and y = r sin θ . Thus z = x + iy = r ( cos θ + i sin θ ) .
1
Clearly r = z = ( x 2 + y 2 ) 2 which is magnitude of the complex number and θ ( undefined if
z = 0 ) is the angle between the positive real axis and the line segment from 0 to z and is
called the argument of z , denoted by θ = arg z .
We note that the value of argument of z is not unique. If θ = arg z , then θ + 2π n ,
where n is an integer is also arg z . The value of arg z that lies in the range −π < θ ≤ π is
called the principal value of arg z .
If z1 , z2 are any two non-zero complex numbers then
1. arg z1 = − arg z1
2. arg z1 z2 = arg z1 + arg z2 .
z 
3. arg  1  = arg z1 − arg z2 .
 z2 
We shall simply state
De Moivre’s Theorem : For any real number n, cos nθ + i sin nθ is one of the values of
(cos θ + i sin θ )n .
4
nth Roots of Complex Numbers.
Let z = r ( cos θ + i sin θ ) be a non-zero complex number, then w = ρ ( cos ϕ + i sin ϕ ) is nth
root of z if wn = z , where n is a positive integer.
Therefore, ρ n ( cos ϕ + i sin ϕ ) = r ( cos θ + i sin θ )
n
ρ n ( cos nϕ + i sin nϕ ) = r ( cos θ + i sin θ )
ρ n = r and nϕ = θ + 2kπ , where k is an integer.
1
Thus
ρ = r n and ϕ =
θ + 2kπ
n
, where k is an integer.
However, only the values of k = 0,1, 2,..., ( n − 1) will give distinct values of w . Hence z
has n distinct nth roots and they are given by
1
  θ + 2 kπ
w = r n cos 
n
 

 θ + 2k π
 + i sin 
n



  where k = 0,1, 2,..., ( n − 1) .

Some Topological aspects
Note that
is a metric space with respect to usual metric d ( z , ζ ) =| z − ζ | .
By an open disc , we mean the set
closed disc, we mean,
{z :
{z :
z − a < ∈} and is denoted by B ( a ; ∈ ) . And by
z − a ≤ ∈} and is denoted by B ( a ; ∈ ) . Further an annulus is
defined as the set {z : r <| z − a |< R} and is denoted by ann(a; r , R ) . The punctured disk of
radius ∈ centered at a is defined by, B ( a ; ∈ ) − {0} = { z : 0 < z − a ≤ ∈} .
Definition 4 A subset G ⊆
is open if, for each z ∈ S , there is an ε > 0 such that
B( z; ε ) ⊆ G . The point z0 is said to be an interior point of the set S ⊆
ε >0
such that
B ( z; ε ) ⊆ S . Further, interior of
S = I{G : G is open and G ⊆ S } . The closure of S ⊆
S = I{F : F is closed and F ⊇ S } . The boundary of
∂S = S ∩ ( X − S ) . Further, a subset S is dense if S =
5
S , written
if there exists an
int S , is the set
, denoted by S , is the set
S , denoted by ∂S and defined by
.
Definition 5 A metric space ( X , d ) is connected if the only subsets of X which are both
open and closed are X and the empty set. Further, a subset S ⊆ X is connected if the metric
space ( S , d ) is connected.
and f : G →
Definition 6 If G is an open set in
in G if lim
h →0
, then f is differentiable at a point a
f ( a + h) − f ( a )
exists. It is denoted by f '(a ) and called derivative of f at a .
h
Definition 7 If f is differentiable on G , then we define f ' : G →
. If f ' is continuous
then we say that f is continuously differentiable.
Definition 8 A differentiable function such that each successive derivative is again
differentiable is called infinitely differentiable.
Definition 9 A function f : G →
is analytic if f is continuously differentiable on G .
Power series
∞
Definition 10 A series of the form
∑ a ( z − a)
n
n
where a1 , a2 ,..., an ,... are constants, is
n =0
called power series about a .
∞
Ex. The geometrical series
∑ z n is power series about 0 and for z < 1 ,
n =0
∞
Theorem 11 For given power series
∑ a ( z − a)
n
n
∞
∑z
n =0
n
=
1
.
1− z
define a number 0 ≤ R ≤ ∞ by
n =0
1
= lim sup an
R
1
n
, then
a) If z − a < R , the series converges absolutely.
b) If z − a > R , the term of series become unbounded and so series diverges.
c) If 0 < r < R , then the series converges uniformly on { z : z − a ≤ r} .
Moreover the number R is the only number having properties (a) and (b).
6
Proof. We may suppose that a = 0 .
a) If
1 1 
z < R , then there is an r such that z < r < R ,  >  . Thus by definition of limit
r R
sup, there is an integer N such that
an
1
n
an <
<
1
for all n ≥ N .
r
1
for all n ≥ N .
rn
 z
an z <  
r 
n
n
n
∞
Thus the series
for all n ≥ N .
∑a z
n
n =0
n
 z
is dominated by the series ∑   . Since the geometric series
r 
n
 z
  converges for z < r , the series
∑
n =0  r 
∞
∞
∑a z
n
n
converges absolutely for each z < R .
n =0
1 1 
b) Suppose z > R , and choose r such that z > r > R ,  <  . Thus by definition of limit
r R
n such that
sup, there are infinitely many integers
 z
an z >  
r 
n
n
an
1
n
>
1
. It follows that
r
 z
for all n ≥ N and , since   > 1 these terms becomes unbounded and so
r 
the series diverges.
c) If 0 < r < R , choose ρ such that r < ρ < R . As in (a) we have an <
n
n
1
ρn
for all n ≥ N .
 z r
r
Thus if z ≤ r , an z ≤   ≤   and   < 1 . Hence, by Weierstrass M-test, the power
ρ
ρ  ρ
n
∞
series
∑a z
n
n
converges uniformly on z ≤ r .
n =0
7
Definition 12 The circle z − a = R which includes in its interior z − a < R , in which the
∞
power series
∑ a ( z − a)
n
converges, is called circle of convergence . Radius R of this
n
n =0
circle is called radius of convergence of power series and in view of above result it is given
by
1
= lim sup an
R
1
n
.
∞
Theorem 13 If
∑ a ( z − a)
n
is given power series with radius of convergence R , then
n
n =0
R = lim
an
if this limit exists.
an +1
Proof. We assume that a = 0 and let α = lim
integer
N
such
r<
that
an
an +1
for
an
. Suppose that z < r < α and find an
an +1
all
n ≥ N.
B = an r N
Let
then
aN +1 r N +1 = aN +1 rr N < aN r N = B ,
aN + 2 r N + 2 = aN + 2 rr N +1 < a N +1 r N +1 < B .
Continuing this way we get, an r n ≤ B for all n ≥ N .
Then an z = an r
n
Since
n
z
n
rn
≤B
z < r we get that
z
n
rn
for all n ≥ N .
∞
∑az
n
n
is dominated by convergent series and hence converges.
n =0
Since r < α was arbitrary this gives that α ≤ R .
Now if z > r > α then an < r an +1 for all n ≥ N . As above we get an r n ≥ B = aN r N for
all n ≥ N . This gives that an z ≥ B
n
z
r
n
n
→ ∞ as n → ∞ . Hence
∞
∑a z
n
n =0
R ≤ α . Thus R = α .
8
n
diverges and so
Example 14 Find radius of convergence for the series
∞
(−1) n
c) ∑
( z − 2i ) n
n
n =0
∞
zn
a) ∑ n
n =0 n
∞
zn
b) ∑
n =0 n !
∞
Solution. a) Here
1
zn
and an = n , a = 0 .
∑
n
n
n =0 n
1
Therefore = lim sup an
R
1
n
1
= lim sup n
n
1
n
= lim sup
1
=0
n
∞
Thus radius of convergence for the series
zn
is R = ∞ .
∑
n
n =0 n
That is the series converges in whole complex plane.
∞
b) Here
zn
∑ n!
n =0
Therefore
and an =
1
, a=0 .
n!
a
n!
1
1
= lim n +1 = lim
= lim
=0
R
an
( n + 1)!
( n + 1)
∞
Thus radius of convergence for the series
zn
is R = ∞ .
∑
n =0 n !
(−1) n
(−1) n
n
c) Here ∑
( z − 2i ) and an =
, a = 2i .
n
n
n =0
∞
Therefore
a
n
1
= lim n +1 = lim
=1
R
an
( n + 1)
(−1) n
( z − 2i ) n
∑
n
n =0
∞
Thus radius of convergence for the series
convergence is z − 2i = 1 .
9
is R = 1 and circle of
∞
Theorem 15 Let f ( z ) = ∑ an ( z − a ) n have radius of convergence R > 0 . Then
n =0
a) For each K ≥ 1 the series
∞
∑ n(n − 1)...(n − k + 1)a ( z − a)
n−k
…(1)
n
has
n= k
radius of convergence R .
b) The function f is infinitely differentiable on B(a, R ) and furthermore, f ( k ) ( z ) is
given by the series (1) for all K ≥ 1 and z − a < R .
c) For n ≥ 0 , an =
1 (n)
f (a) .
n!
Proof. With no loss of generality assume that a = 0 .
∞
Therefore
f ( z ) = ∑ an z n
…(2)
n =0
a) We first prove the result for K = 1 . That is the power series
∞
∑ an z n and
n =0
∞
∑ na z
n
n =1
same radius of convergence.
∞
Let g ( z ) = ∑ nan z n −1 have radius of convergence R ' where
n =1
∞
Since R is radius of convergence of f ( z ) = ∑ an z n ,
n =0
1
= lim sup nan
R'
1
n −1
1
1
= lim sup an n .
R
Now we have to show that R = R ' .
1
1
1




log n
log  lim n n −1  = lim  log n n −1  = lim
= lim n = 0
n −1
1




1
Therefore lim n n −1 = e0 = 1 .
Thus lim sup nan
1
n −1
1


=  lim n n −1   lim sup an



= 1.  lim sup an

∞
Therefore the series
∑ nan z n−1 and
n =1
If z < R ' , we write
∞
∑
n =0
1
n −1
1
n −1




 = lim sup an

∞
∑a z
n −1
n
1
n −1
have same radius of convergence.
n =1
an z n = a0 + z
∞
∑a z
n
n =1
10
n −1
< ∞.
n −1
have
∞
∑az
That is if z < R ' ,
is convergent. Hence R ≥ R ' .
n
n
n =0
∞
∑az
Also if z < R , we write
n −1
n
1
z
=
n =1
∞
∑a z
That is if z < R ,
n −1
∞
∑a z
n
n
−
a0
n=0
≤
z
1
z
∞
∑az
n
n =0
n
+
a0
z
< ∞ for z ≠ 0 .
is convergent. Hence R ≤ R ' .
n
n =1
Thus R = R ' .
∞
Thus
∑ an z n and
n =0
∞
∑ na z
∑ nan z n−1 and
n =1
Therefore
have same radius of convergence.
n =1
∞
Similarly
n −1
n
by
∞
∑ n(n − 1)a z
n−2
n
have same radius of convergence R .
n= 2
method
∞
∑ n(n − 1)...(n − k + 1)a ( z − a)
of
n−k
n
induction
for
any
K ≥1
the
has radius of convergence R .
n= k
n
b) For z < R , let S n ( z ) = ∑ ak z k and Rn ( z ) =
k =0
∞
∑az
k
k
so that f ( z ) = S n ( z ) + Rn ( z ) .
k = n +1
Now fix a point w ∈ B ( 0; R ) , then there is 0 < r < R such that w < r < R .
Let δ > 0 be such that B ( w; δ ) ⊆ B ( 0; r ) . Let z ∈ B ( w; δ ) .
Consider
S ( z ) − S n ( w)
R ( z ) − Rn ( w)
f ( z ) − f ( w)
− g ( w) = n
− S n '( w) + S n '( w) − g ( w) + n
z−w
z−w
z−w
And
∞
Rn ( z ) − Rn ( w)
=
z−w
=
∑ a (z
k
k = n +1
k
− wk )
z−w
∞
∑
≤
∞
∑
k = n +1
ak
z k − wk
z−w
ak z k −1 + z k − 2 w + ... + z wk − 2 + wk −1
k = n +1
≤
∞
∑
k = n +1
ak
{z
k −1
+z
k −2
w + ... + z w
11
k −2
+w
k −1
}
…(3)
series
∞
∑
≤
k = n +1
∞
=
∑
ak {r k −1 + r k − 2 r + ... + r r k − 2 + r k −1}
ak k r k −1
k = n +1
Since r < R ,
∞
∑
ak k r k −1 converges.
k = n +1
Therefore for any ∈> 0 , there is an integer N1 > 0 such that
∞
∑
ak k r k −1 < ∈
k = n +1
3
whenever
n ≥ N1 .
Thus for n ≥ N1
Rn ( z ) − Rn ( w) ∈
<
3
z−w
…(4)
Since lim S n '( w) = g ( w) , there is an integer N 2 > 0 such that S n '( w) − g ( w) < ∈
n →∞
3
whenever n ≥ N 2 .
Let n = max { N1 , N 2 } .
S n ( z ) − S n ( w)
= S n '( w) , for given ∈> 0 , we choose δ > 0 such that
z →w
z−w
Since lim
S n ( z ) − S n ( w)
− S n '( w) < ∈
3
z−w
whenever 0 < z − w < δ .
Thus for given ∈> 0 , there is δ > 0 such that
f ( z ) − f ( w)
− g ( w) < ∈ + ∈ + ∈ =∈
3
3
3
z−w
whenever 0 < z − w < δ .
Hence f is differentiable and f '( w) = g ( w) for all w ∈ B ( 0; R ) .
12
…(6)
…(5)
∞
That is f '( z ) = ∑ nan z n −1 .
n =1
∞
Similarly f ''( z ) = ∑ n(n − 1)an z n − 2
n=2
∞
f '''( z ) = ∑ n(n − 1)(n − 2)an z n −3
n =3
∞
and so on f ( k ) ( z ) = ∑ n(n − 1)(n − 1)...(n − k + 1)an z n − k .
n=k
c) From part (b) we have
f (0) = a0 , f ''(0) = 1. a1 , f '''(0) = 1.2. a2 ,…, f ( k ) (0) = 1.2.3....k . a2 .
Thus ak =
1 (k )
f (0)
k!
∞
Hence for f ( z ) = ∑ an ( z − a ) n ,
an =
n =0
1 (n)
f (a) .
n!
∞
Corollary 16 If the series
∑ a ( z − a)
has radius of convergence R > 0 then,
n
n
n =0
∞
f ( z ) = ∑ an ( z − a ) n is analytic in B(a, R ) .
n =0
∞
Proof. By above theorem if
∑ a ( z − a)
n
n
has radius of convergence R > 0 , then
n =0
∞
f ( z ) = ∑ an ( z − a ) n is infinitely differentiable in B(a, R ) .
n =0
Therefore f ' , f '' exists in B(a, R ) implies that f ' is continuous in B(a, R ) .
Thus f is continuously differentiable. Hence f is analytic in B(a, R ) .
Result 17 A domain G is connected iff its open as well as closed subset is either empty
or G .
13
Theorem 18 If G is open and connected and f : G →
is differentiable with f '( z ) = 0 for
all z in G , then f is constant.
Proof. Fix z0 in G and let w0 = f ( z0 ) . Let A = { z ∈ G : f ( z ) = w0 } . Clearly A ≠ .
If A is open as well as closed, then by connectedness of G , A = G . ( i.e. f is constant )
First we prove that A - is open.
Now for a ∈ A . Let ∈> 0 be such that B ( a;∈) ⊂ G .
If z ∈ B ( a;∈) we define g :[0,1] → G by g (t ) = f [ t z + (1 − t ) a ] , 0 ≤ t ≤ 1 .
Then
g (t ) − g ( s ) f [ t z + (1 − t ) a ] − f [ s z + (1 − s ) a
=
t−s
t−s
=
=
lim
t→s
f [ t z + (1 − t ) a ] − f [ s z + (1 − s ) a
[t z + (1 − t ) a ] − [ s z + (1 − s) a ]
f [ t z + (1 − t ) a ] − f [ s z + (1 − s ) a
[t z + (1 − t ) a ] − [ s z + (1 − s) a ]
]
] . [t z + (1 − t ) a ] − [ s z + (1 − s) a ]
t−s
] .( z − a)
f [ t z + (1 − t ) a ] − f [ s z + (1 − s ) a ]
g (t ) − g ( s )
= lim
.( z − a )
t→s
t−s
[t z + (1 − t ) a ] − [ s z + (1 − s) a ]
g '( s ) = f ' [ s z + (1 − s ) a ] .( z − a) = 0.( z − a ) = 0
Therefore g '( s ) = 0 for 0 ≤ s ≤ 1 implies that g is constant.
Hence g (1) = g (0) implies that f ( z ) = f (a ) = w0 .Therefore z ∈ A .
Thus if z ∈ B ( a;∈) then z ∈ A that is B ( a;∈) ⊂ A . Thus A is open.
We now prove that A - is closed.
Let z be limit point of A , then there is a sequence
(
)
{ zn }
in A such that lim zn = z .
n →∞
Since f is continuous , f ( z ) = f lim zn = lim f ( zn ) = lim w0 = w0 . Hence z ∈ A .
n →∞
n →∞
Thus A contains all its limit points hence A is closed.
14
n →∞
EXERCISES
1) Find the radius of convergence of the followings
1
1
a) ∑ n z n
b) ∑ z n
n
n!
∞
1 n
z
e) ∑ a n z n a ∈
d) ∑
2
1 + in
n=0
c)
∑ (3 + 4i)
∞
f)
∑ k n zn
n
zn
k∈ .
n=0
15
UNIT - II
MÖBIUS TRANSFORMATIONS
In this unit we study Möbius transformations and their properties. We begin with bilinear
transformation.
Definition 19 A mapping of the form S ( z ) =
transformation where a, b, c, d ∈
az + b
is called bilinear or linear fractional
cz + d
.
Definition 20 A bilinear transformation S ( z ) =
az + b
with ad − bc ≠ 0 is called Möbius
cz + d
map or Möbius transformation.
Remarks 21 1) Möbius transformation is one-one and onto.
2) If S ( z ) =
az + b
− dw + b
, then S −1 ( w) =
.
cz + d
cw − a
3) If S and T are Möbius transformations then S o T is also Möbius
transformation.
4) S ( z ) = z + a ( Translation )
S ( z ) = az
( Dilation/Magnification )
S ( z ) = eiθ z ( Rotation )
S ( z) =
1
z
( Inversion ).
Theorem 22 If S is a Möbius transformation then S is composition of translation , dilation
and inversion.
16
Proof. Let S ( z ) =
az + b
with ad − bc ≠ 0 be Möbius transformation.
cz + d
a b
Case 1. When c = 0 then S ( z ) =   z +  
d  d 
a
b
Let S1 ( z ) =   z , S 2 ( z ) = z +  
d 
d 
 a    a   b 
Then S 2 o S1 ( z ) = S 2 ( S1 ( z ) ) = S 2    z  =   z +   = S ( z )
 d    d   d 
Thus S = S 2 o S1 .
Case 2. When c ≠ 0
Let S1 ( z ) = z +
d
1
bc − ad
a
, S 2 ( z ) = , S3 ( z ) =
z , S4 ( z ) = z + .
2
c
z
c
c
Then S4 o S3 o S2 o S1 ( z ) = S4 o S3 o S2 ( S1 ( z ) )
d

= S 4 o S3 o S 2  z + 
c

 
d 
= S 4 o S3  S 2  z +  
c 
 

 1
= S 4 o S3 
z+d
c






 
  1
= S 4  S3 
  z + d
c
 



 



 bc − ad  1
= S4 

2
d
 c
z+
c





 

 bc − ad  a
=
+
 c(cz + d )  c
17
=
az + b
= S ( z) .
cz + d
Thus S = S 4 o S3 o S 2 o S1 .
Theorem 23 Every Möbius transformation can have at most two fixed points.
Proof. Let S ( z ) =
az + b
with ad − bc ≠ 0 be Möbius transformation.
cz + d
Let z be fixed point of S ( z ) then S ( z ) = z
az + b
=z
cz + d
cz 2 + (d − a ) z − b = 0
which is quadratic in z . Hence it can have at most two roots. Therefore every Möbius
transformation can have at most two fixed points otherwise S ( z ) = z for all z (Identity map ).
Theorem 24 Möbius map is uniquely determined by its action on any three distinct points in
∞
.
Proof. Let z1 , z2 , z3 be three distinct points in
∞
. Let S and T be Möbius map such that
S ( z1 ) = w1 , S ( z2 ) = w2 , S ( z3 ) = w3 and T ( z1 ) = w1 , T ( z2 ) = w2 , T ( z3 ) = w3 .
Then
T −1 o S ( z1 ) = T −1 ( S ( z1 ) ) = T −1 ( w1 ) = z1
T −1 o S ( z2 ) = T −1 ( S ( z2 ) ) = T −1 ( w2 ) = z2
T −1 o S ( z3 ) = T −1 ( S ( z3 ) ) = T −1 ( w3 ) = z3
Thus T −1 o S is a Möbius map having three fixed points.
Hence T −1 o S = I ( Identity map). Thus T = S .
Definition 25 For z ∈
∞
the map denoted by ( z , z2 , z3 , z4 ) where z2 , z3 , z4 ∈
cross ratio if it maps z2 , z3 , z4 respectively to 1, 0, ∞ .
18
∞
is called
More precisely the map z a ( z , z2 , z3 , z4 ) is a Möbius map that maps z2 , z3 , z4 respectively
to 1, 0, ∞ and is given by
 z − z3   z 2 − z 3 
S ( z ) = ( z , z 2 , z3 , z 4 ) = 
 
.
 z − z 4   z2 − z4 
Remarks 26 1) ( z2 , z2 , z3 , z4 ) = 1, ( z3 , z2 , z3 , z4 ) = 0, ( z4 , z2 , z3 , z4 ) = ∞ .
2) ( z ,1, 0, ∞ ) = z .
M ( w2 ) = 1, M ( w3 ) = 0, M ( w4 ) = ∞
3) Let M be any Möbius map such that
then M ( z ) = ( z , w2 , w3 , w4 ) .
Theorem 27 If
z 2 , z3 , z 4
are distinct points and T
( z1 , z2 , z3 , z4 ) = (Tz1 , Tz2 , Tz3 , Tz4 )
is any Möbius map then
for any point z1 . ( Cross-ratio is invariant under any
Möbius map )
Proof. Let S ( z ) = ( z , z2 , z3 , z4 ) and T is any Möbius map. Let M = S o T −1 , then M o T = S .
Now,
1 = S ( z 2 ) = M o T ( z 2 ) = M ( T ( z2 ) )
0 = S ( z3 ) = M o T ( z 3 ) = M ( T ( z 3 ) )
∞ = S ( z4 ) = M o T ( z4 ) = M ( T ( z 4 ) )
Thus M ( z ) = ( z , Tz2 , Tz3 , Tz4 ) then S o T −1 ( z ) = ( z , Tz2 , Tz3 , Tz4 ) .
Let T −1 ( z ) = z1 then z = T ( z1 ) . Therefore S ( z1 ) = (Tz1 , Tz2 , Tz3 , Tz4 ) .
Thus ( z1 , z2 , z3 , z4 ) = (Tz1 , Tz2 , Tz3 , Tz4 ) .
Theorem 28 If z2 , z3 , z4 are distinct points in
∞
∞
and w2 , w3 , w4 are also distinct points of
, then is one and only one Möbius map S such that S ( z2 ) = w2 , S ( z3 ) = w3 , S ( z4 ) = w4 .
Proof. Let T ( z ) = ( z , z2 , z3 , z4 ) and M ( z ) = ( z , w2 , w3 , w4 ) . Let S = M −1 o T then
S ( z2 ) = M −1 o T ( z2 ) = M −1 (T ( z2 ) ) = M −1 (1) = w2
S ( z3 ) = M −1 o T ( z3 ) = M −1 ( T ( z3 ) ) = M −1 ( 0 ) = w3
S ( z4 ) = M −1 o T ( z4 ) = M −1 (T ( z4 ) ) = M −1 ( ∞ ) = w4
19
Thus we have a Möbius map S such that S ( z2 ) = w2 , S ( z3 ) = w3 , S ( z4 ) = w4 .
Uniqueness:
Let R be another Möbius map such that R ( z2 ) = w2 , R ( z3 ) = w3 , R ( z4 ) = w4 .
Then R −1 o S ( z2 ) = R −1 ( S ( z2 ) ) = R −1 ( w2 ) = z2
R −1 o S ( z3 ) = R −1 ( S ( z3 ) ) = R −1 ( w3 ) = z3
R −1 o S ( z4 ) = R −1 ( S ( z4 ) ) = R −1 ( w4 ) = z4
Thus R −1 o S has three fixed points z2 , z3 , z4 implies that R −1 o S = I .
Therefore, R = S .
Example 29 Evaluate following cross ratios a)
( 7 + i,1, 0, ∞ )
b) ( 2,1 − i,1,1 + i )
 z − z3   z 2 − z 3 
Sol. We have, S ( z ) = ( z , z2 , z3 , z4 ) = 
 

 z − z 4   z2 − z4 
 z − z3 
a) ( z , z2 , z3 , ∞ ) = 

 z 2 − z3 
Therefore,
7+i−0
 = 7+i.
 1− 0 
( 7 + i,1, 0, ∞ ) = 
 2 −1 
b) ( 2,1 − i,1,1 + i ) = 

 2 − (1 + i ) 
 (1 − i ) − 1   1   −i 
2
= 1+ i .

=
 
=
 (1 − i ) − (1 + i )   1 − i   −2i  1 − i
Example 30 Find Möbius map which maps the points z2 = 2, z3 = i, z4 = −2 onto
w2 = 1, w3 = i, w4 = −1 respectively.
Solution Let S be the map that takes zi a wi
( i = 2,3, 4 ) .
under any Möbius map, ( z , z2 , z3 , z4 ) = ( Sz , Sz2 , Sz3 , Sz4 ) .
Therefore,
Thus
( z, z2 , z3 , z4 ) = ( w, w2 , w3 , w4 )
, where S ( z ) = w .
( z, 2, i, −2 ) = ( w,1, i, −1)
 z − i   2 − i   w − i   1− i 

 
=
 

 z + 2   2 + 2   w +1   1+1 
20
Since cross ratio is invariant
4( z − i)
=
2(w − i)
( z + 2 )( 2 − i ) ( w + 1)(1 − i )
2( z − i)
2 z − iz + 4 − 2i
=
(w − i)
w − iw + 1 − i
2 ( z − i )( w − iw + 1 − i ) = ( w − i )( 2 z − iz + 4 − 2i )
w ( 2 ( z − i ) − 2 ( z − i ) i ) + 2 ( z − i )(1 − i ) = w ( 2 z − iz + 4 − 2i ) − i ( 2 z − iz + 4 − 2i )
w=
−2 ( z − i )(1 − i ) − i ( 2 z − iz + 4 − 2i )
( −2 (1 − i ) − i ( 2 − i ) ) z + ( 2i (1 − i ) − i ( 4 − 2i ) )
=
2 ( z − i ) − 2 ( z − i ) i − ( 2 z − iz + 4 − 2i )
( 2 − 2i − ( 2 − i ) ) z + ( −2i + 2i 2 − ( 4 − 2i ) )
Therefore, w =
( −3) z + ( −2i ) = 3 z + 2i .
( −i ) z + ( −6 ) iz + 6
Theorem 31 Let z1 , z2 , z3 , z4 be four distinct points in
∞
, then ( z1 , z2 , z3 , z4 ) is real number
iff all four points lie on the circle.
Proof. Let S ( z ) = ( z , z2 , z3 , z4 ) then S is a Möbius map from
theorem we have to prove that
{ w∈
∞
to
∞
. To prove this
: S ( w) = real } is a circle.
Suppose
S ( w) = real , then S ( w) = S ( w) .
Let S ( w) =
aw + b
with ad − bc ≠ 0 .
cw + d
Thus,
∞
aw + b aw + b
=
cw + d cw + d
2
Therefore, (ac − ac) w + (ad − bc) w + (bc − ad ) w + (bd − bd ) = 0
…(1)
Case 1. When ac is real.
Therefore, ac = ac , then from (1) we have,
(ad − bc) w + (bc − ad ) w + (bd − bd ) = 0
Let α = 2(ad − bc), β = i (bd − bd ) then (2) becomes,
21
…(2)
α
2
w+
α
2
w+
(
β
=0
i
)
i α w + α w + 2β = 0
i.2i.Im (α w ) + 2β = 0
Im (α w ) − β = 0
…(3)
Let α = p + iq, w = x + iy then α w = px − qy + i (qx + py ) .
 −q 
Therefore, Im (α w ) − β = (qx + py ) − β = 0 . Thus (3) represents a line y =   x + β .
 p 
That is, w lies on the line determined by (3) for fixed α and β . We know that straight line
may be regarded as circle with infinite radius. Therefore, w lies on the circle.
Case 2. When ac is not real.
Therefore, ac ≠ ac , then from (1) we have,
2
w +
(ad − bc)
(bc − ad )
(bd − bd )
w+
w+
=0
(ac − ac)
(ac − ac)
(ac − ac)
 ad − bc 
 bd − bd 
Let γ = 
, δ = −
.
ac
−
ac
ac
−
ac




2
Therefore, w + γ w + γ w − δ = 0
ww + γ w + γ w + γ γ = δ + γ γ
(w + γ )(w + γ ) = δ + γ γ
2
w+γ =δ +γγ
2
w+γ =δ +γγ
Therefore, w + γ = λ
(
where λ = δ + γ γ
)
1
2
…(4)
=
ad − bc
>0
ac − ac
Since γ and λ are independent of w , (4) represents a circle on which w lies.
22
Theorem 32 A Möbius transformation takes circles onto circles.
Proof. Let S be a Möbius transformation. Let Γ be a circle in
points on
∞
Γ such that S ( z2 ) = w2 , S ( z3 ) = w3 , S ( z4 ) = w4 . Then
and z2 , z3 , z4 are distinct
w2 , w3 , w4
determine a
circle Γ ' .
We claim that S ( Γ ) = Γ ' :
Since cross ratio is invariant under any Möbius transformation, for any z in
∞
,
( z, z2 , z3 , z4 ) = ( Sz, Sz2 , Sz3 , Sz4 )
= ( Sz, w2 , w3 , w4 )
Now z ∈ Γ ⇔
⇔
( z, z2 , z3 , z4 )
is real.
( Sz, w2 , w3 , w4 ) is real.
⇔ S ( z)∈Γ'
Thus S ( Γ ) = Γ ' .
Theorem 33 For any given circles Γ and Γ ' in
∞
there is a Möbius transformation T such
that T ( Γ ) = Γ ' . Furthermore we can specify that T takes any three points on Γ onto any
three points of Γ ' . If we do specify T ( z j ) = w j for j = 2, 3, 4 (distinct z j in Γ ) then T is
unique.
Proof. Let z2 , z3 , z4 be distinct points on Γ and w2 , w3 , w4 be points on Γ ' . Let
S ( z ) = ( z, z2 , z3 , z4 ) and M ( z ) = ( z, w2 , w3 , w4 ) .
Let T = M −1 o S , then T ( z2 ) = M −1 o S ( z2 ) = M −1 ( S ( z2 ) ) = M −1 (1) = w2
T ( z3 ) = M −1 o S ( z3 ) = M −1 ( S ( z3 ) ) = M −1 ( 0 ) = w3
T ( z4 ) = M −1 o S ( z4 ) = M −1 ( S ( z4 ) ) = M −1 ( ∞ ) = w4
Thus T is a Möbius transformation that takes Γ onto Γ ' .
Obviously Möbius transformation is unique.
23
EXERCISES
1) Find fixed points of dilation, translation and inversion on C∞.
2) If Tz =
αz + β
az + b
and Sz =
. Prove that T = S iff α = aλ , β = bλ γ = cλ , δ = d λ ,
γ z +δ
cz + d
for some complex number λ.
24
UNIT - III
COMPLEX INTEGRATION
In this section we shall study complex integrations of complex functions and
established fundamental theorem of calculus for line integral. We show that an analytic
function has a power series expansion
as a Taylor’s theorem. Form then we established
Cauchy’s estimate to prove Cauchy’s theorem. We begin with elementary definitions
Definition 1 A path in region G ⊂
interval [a, b] in
is a continuous function
γ :[a, b] → G for some
.
If γ '(t ) exists for each t in [a, b] and γ ' :[a, b] →
is continuous then γ is called smooth
path. Also γ is called piecewise smooth if there is partition of [a, b] , a = t0 < t1 < ... < tn = b ,
such that γ is smooth on each subinterval [t j −1 , t j ], 1 ≤ j ≤ n .
Definition 2 Let
γ :[a, b] → G be a path then trace of γ is
{ γ (t ) : t ∈ [a, b] }
and it is
denoted by {γ } .
i.e. {γ } = { γ (t ) : t ∈ [a, b] } . Note that trace of a path is always compact.
Definition 3 A function γ :[a, b] →
, for [a, b] ⊂
, is of bounded variation if there is a
constant M > 0 such that for any partition P = {a = t0 < t1 < ... < tm = b} of [a, b]
m
υ ( γ ; P ) = ∑ γ (tk ) − γ (tk −1 ) ≤ M .
k =1
The
total
variation
of
γ,
denoted
by
V (γ )
is
defined
V ( γ ) = sup {υ ( γ ; P ) : P a partition of [a,b] } .
Definition 4 A path γ :[a, b] →
is rectifiable if γ is a function of bounded variation.
25
as
Theorem 5 Let γ :[a, b] →
be of bounded variation. Then:
a) If P and Q are partitions of [a, b] and P ⊂ Q then υ ( γ ; P ) ≤ υ ( γ ; Q ) .
b) If σ :[a, b] →
is also of bounded variation and α , β ∈
then αγ + βσ is of bounded
variation and V (αγ + βσ ) ≤ α V ( γ ) + β V (σ ) .
If γ :[a, b] →
Theorem 6
is piecewise smooth then γ is of bounded variation and
b
V ( γ ) = ∫ γ '(t ) dt .
a
Proof. Firstly we assume that
γ
is smooth so that
γ ' is continuous. Let
P = {a = t0 < t1 < ... < tm = b} then
m
m
υ ( γ ; P ) = ∑ γ (tk ) − γ (tk −1 ) = ∑
k =1
m
tk
∫ γ '(t )dt
k =1 tk −1
≤∑
tk
∫
γ '(t ) dt
k =1 tk −1
b
= ∫ γ '(t ) dt .
a
b
Hence, V ( γ ) ≤ ∫ γ '(t ) dt , so that γ is of bounded variation.
a
Since γ ' is continuous it is uniformly continuous. Thus for given ∈> 0 we can choose δ1 > 0
such that
γ '( s) − γ '(t ) < ∈ whenever s − t < δ1 . Also we choose δ 2 > 0 such that if
P = {a = t0 < t1 < ... < tm = b} and P = max {( tk − tk −1 ) :1 ≤ k ≤ m} < δ 2 then
b
m
tk
∫ γ '(t ) dt − ∑ ∫
a
γ '(τ k ) ( tk − tk −1 ) < ∈ where τ k is any point in [tk −1 , tk ] .
k =1 tk −1
b
m
tk
∫ γ '(t ) dt ≤ ∈ + ∑ ∫
a
γ '(τ k ) ( tk − tk −1 )
k =1 tk −1
m
=∈+ ∑
tk
∫ γ '(τ
k
)dt
k =1 tk −1
26
m
≤∈+ ∑
tk
m
∫ [γ '(τ k ) − γ '(t )] dt + ∑
k =1 tk −1
tk
∫ γ '(t )dt
k =1 tk −1
If P < δ = min (δ1,δ 2 ) then γ '(τ k ) − γ '(t ) <∈ for t in [tk −1 , tk ] and
b
m
a
k =1
∫ γ '(t ) dt ≤ ∈ + ∈ ( b − a ) + ∑ γ (tk ) − γ (tk −1 )
= ∈ 1 + ( b − a )  + υ ( γ ; P )
≤ ∈ 1 + ( b − a )  + V ( γ ) .
b
∫ γ '(t ) dt ≤ V (γ ) .
Letting ∈→ 0 + we get
a
b
Thus V ( γ ) = ∫ γ '(t ) dt .
a
Theorem 7 Let f and g be continuous functions on [a, b] and let γ and σ be functions of
bounded variation on [a, b] . Then for any scalar α and β :
a)
b
b
b
a
a
a
b
b
a
a
∫ (α f + β g ) d γ = α ∫ fd γ + β ∫ gdγ
b
b)
∫
a
fd (αγ + βσ ) = α ∫ fd γ + β ∫ gdσ .
Theorem 8 If
b
∫
a
γ
is piecewise smooth and
f :[a, b] →
is continuous then
b
f d γ = ∫ f (t ) γ '(t )dt .
a
Proof. To prove this theorem we consider real and imaginary parts of γ , we reduce the proof
to the case where
γ ( [ a, b] ) ⊂
. For any ∈> 0
P = {a = t0 < t1 < ... < tn = b} has P < δ then
b
∫f
a
n
dγ − ∑
tk
∫
k =1 tk −1
f (τ k ) γ ( tk ) − γ ( tk −1 )  <
∈
and
2
27
choose
δ >0
such that if
b
∫
n
f (t )γ '(t )dt − ∑
tk
∫
f (τ k )γ '(τ k ) ( tk − tk −1 ) <
k =1 tk −1
a
∈
for any choice of τ k in [tk −1 , tk ]
2
Now by Mean Value Theorem
n
∑
k =1
n
f (τ k ) γ ( tk ) − γ ( tk −1 )  = ∑ f (τ k )γ '(τ k ) ( tk − tk −1 ) for some τ k in [tk −1 , tk ] .
k =1
Combining this with above two inequalities we get
b
∫
a
b
f d γ − ∫ f (t ) γ '(t )dt < ∈. Since ∈> 0 was arbitrary we have,
a
Definition 9 If γ :[a, b] →
b
∫
a
b
f d γ = ∫ f (t ) γ '(t )dt .
a
is a rectifiable path and f is a function defined and continuous
b
on the trace of γ then (line) integral of f along γ is
∫ f (γ (t ) ) d γ ( t ) . This line integral is
a
also denoted by
∫γ f = ∫γ f ( z)dz .
Definition 10 Let γ :[a, b] →
and σ :[c, d ] →
be rectifiable paths. The path σ is
equivalent to γ if there is a continuous function ϕ :[c, d ] → [a, b] , which is strictly
increasing, and with ϕ (c) = a, ϕ (d ) = b , such that σ = γ o ϕ .
The idea is to recognize all the paths having same trace as identical. The above definition
brings about a partition of the class of all paths. Thus we are prompted to define.
A curve is an equivalence class of paths. The trace of a curve is the trace of any one of its
members. A curve is smooth (piecewise smooth) if and only if one of its representative is
smooth (piecewise smooth). A curve C is called simple if it does not cross over itself. That is
γ ( t1 ) ≠ γ ( t2 ) whenever t1 ≠ t2 . A curve C is called a simple closed curve if i) γ (a) = γ (b)
ii) γ ( t1 ) ≠ γ ( t2 ) whenever t1 ≠ t2 , except when t1 = a and t2 = b . It is also called Jordan
curve.
28
Theorem 11 Let γ be a rectifiable curve and suppose that f is a function continuous on
{γ } . Then:
∫γ f
a)
b)
= −∫ f ;
−γ
∫γ f ≤ ∫γ
c) If c ∈
f dz ≤ V (γ ) sup  f ( z ) : z ∈ {γ } ;
then
∫γ f ( z)dz = γ ∫
f ( z − c)dz .
+c
We shall conclude this chapter with
Theorem 12 Let G be an open set in
and let γ be a rectifiable path in G with initial and
end points α and β respectively. If f : G →
F :G →
, then
∫γ f
is a continuous function with primitive
= F ( β ) − F (α ) .
We now prove Leibnitz theorem:
Theorem 13 Let ϕ :[a, b] × [c, d ] →
be a continuous function. Define g :[c, d ] →
by
b
g (t ) = ∫ ϕ ( s, t ) ds , ∀ t ∈ [c, d ]
a
Then,
i)
g is continuous function and
ii)
If
∂ϕ
exists and continuous, then g is continuously differentiable.
∂t
∂
ϕ ( s, t ) ds .
∂t
a
b
Moreover, g '(t ) = ∫
Proof : i) Let t0 ∈ [c, d ] and ∈> 0 .
Since ϕ is continuous on [a, b] × [c, d ] , we have ϕ is uniformly continuous on
[ a , b ] × [c, d ] .
Therefore, there is δ > 0 such that for each s ∈ [a, b] , we have,
29
ϕ ( s , t 0 + h ) − ϕ ( s , t0 ) <
∈
, whenever t0 + h − t0 < δ
b−a
Now, for h < δ we have
b
b
∫ ϕ ( s, t0 + h) ds − ∫ ϕ ( s, t0 ) ds
g (t0 + h) − g (t0 ) =
a
a
b
≤
∫
b
ϕ ( s, t0 + h) − ϕ ( s, t0 ) ds ≤
a
∫
a
b
=
∫ [ ϕ ( s, t
0
+ h) − ϕ ( s, t0 ) ] ds
a
∈
∈
ds =
(b − a )
b−a
b−a
=∈
g (t0 + h) − g (t0 ) ≤ ∈ , whenever h < δ
Thus,
i.e. g is continuous at t0 ∈ [c, d ]
Since t0 is arbitrary element of [c, d ] , we have g is continuous on [c, d ] .
ii) Suppose that
∂ϕ
exists and continuous
∂t
Let t0 ∈ [c, d ] and ∈> 0
∂
ϕ ( s, t )
∂t
Denote, ϕ 2 ( s, t ) =
Again since ϕ 2 ( s, t ) is uniformly continuous on [a, b] × [c, d ] , ∃ δ > 0 such that for each
s ∈ [a, b] , we have
ϕ2 ( s, t ) − ϕ2 ( s, t0 ) < ∈ , whenever t − t0 < δ
t
Thus, for t − t0 < δ and s ∈ [a, b] , we have
∫ [ ϕ2 (s,τ ) − ϕ2 ( s, t0 ) ] dτ
t0
t
< ∫ ∈ dτ
t0
t
i.e
∫ [ ϕ (s,τ ) − ϕ (s, t ) ] dτ
2
2
0
< ∈ t − t0 , whenever t − t0 < δ and s ∈ [a, b]
t0
… (1)
Let Φ(t ) = ϕ ( s, t ) − t ϕ2 ( s, t0 ) , for some fixed s ∈ [a, b] . Then Φ '(t ) = ϕ2 ( s, t ) − ϕ2 ( s, t0 )
i.e. Φ(t ) is primitive of ϕ 2 ( s, t ) − ϕ2 ( s, t0 )
30
Then by Fundamental Theorem of Calculus for Line Integrals, we have
t
∫ [ ϕ ( s,τ ) − ϕ (s, t ) ] dτ
2
2
0
= Φ (t ) − Φ(t0 )
t0
= [ϕ ( s, t ) − ϕ ( s, t0 ) ] − ( t − t0 ) ϕ2 ( s, t0 )
Inequality (1) becomes,
[ϕ ( s, t ) − ϕ ( s, t0 )] − ( t − t0 ) ϕ2 ( s, t0 )
< ∈ t − t0
…(2)
Now,
b
b
b
 b
g (t ) − g (t0 )
1 
− ∫ ϕ2 ( s, t0 ) ds =
ϕ
(
s
,
t
)
ds
−
ϕ
(
s
,
t
)
ds

 − ∫ ϕ2 ( s, t0 ) ds
0
∫a
t − t0
t − t0  ∫a
a
 a
 ϕ ( s, t ) − ϕ ( s , t0 ) − ( t − t0 ) ϕ 2 ( s , t0 ) 
 ds
t − t0

a
b
∫ 
=
b
≤
∫
ϕ ( s, t ) − ϕ ( s , t0 ) − ( t − t0 ) ϕ 2 ( s , t0 )
t − t0
a
ds
b
< ∈ ∫ ds = ∈ ( b − a )
(By equation (2))
a
Thus, for each ∈> 0 , ∃ δ > 0 such that
g (t ) − g (t0 ) b
− ∫ ϕ2 ( s, t0 ) ds < ∈ ( b − a ) , whenever 0 < t − t0 < δ
t − t0
a
i.e. g is differentiable at t0 and hence on [c, d].
b
∂
ϕ ( s, t0 )ds
∂t
a
b
Next, g '(t0 ) = ∫ ϕ2 ( s, t0 ) ds = ∫
a
As ϕ 2 ( s, t ) =
∂ϕ
is continuous, we have g ' is continuous
∂t
Hence, g is continuously differentiable.
31
2π
∫
Example 14 Prove that
0
eis
ds = 2π , when z < 1
eis − z
2π
Solution: Define g :[0,1] →
by g (t ) =
∫
0
eis
ds , ∀ t ∈ [0,1]
eis − tz
By Leibnitz rule,
2π
g '(t0 ) =
∫
0
2π
∂  eis 
z eis
ds
=


∫0 eis − tz 2 ds
∂t  eis − tz 
(
)
Let Φ( s ) =
zi
z eis
s
.
Then
Φ
'(
)
=
2
eis − tz
( eis − tz )
Thus, g '(t ) = Φ (2π ) − Φ(0)
=
(By fundamental theorem of calculus for line integrals)
zi
zi
− 0i
e − tz e − tz
2π i
=0
i.e. g '(t ) = 0, ∀ t ∈ [0,1]
Therefore, g is constant function
In particular, we have g (1) = g (0)
2π
i.e.
∫
0
2π
i.e.
∫
0
eis
ds =
eis − z
2π
∫
ds
0
eis
ds = 2π .
eis − z
Exercise 15 Show that
z
dz
= 2π i, when a < 1.
z −a
=1
∫
(Hint: put z = eit and use above example) .
The following theorem is known as Cauchy Integral Formula.
32
be analytic and suppose B(a, r ) ⊂ G (r > 0) . If γ (t ) = a + reit ,
Theorem 16 Let f : G →
0 ≤ t ≤ 2π , then f ( z ) =
1
f ( w)
dw , for z − a < r .
∫
2π i γ w − z
Proof. Without loss of generality assume a = 0 and r = 1
That is we may assume that B(0,1) ⊂ G
Let z ∈ B(0,1) ; we have to show that
f ( z) =
1
f ( w)
dw
∫
2π i γ w − z
1
=
2π i
2π
i.e. 2π f ( z ) =
∫
0
2π
i.e.
∫
0
2π
∫
0
f (eis )eis i
ds
eis − z
f (eis )eis
ds
eis − z
 f (eis )eis

 eis − z − f ( z ) ds = 0


Let ϕ ( s, t ) =
f ( z + t (eis − z ))eis
− f ( z ) where 0 ≤ s ≤ 2π and 0 ≤ t ≤ 1
eis − z
z + t (eis − z ) =
Since
z (1 − t ) + teis ≤ (1 − t ) z + t eis < 1 , ϕ
continuously differentiable
Now define g :[0,1] →
by
2π
g (t ) =
∫ ϕ ( s, t ) ds
0
Then by Leibnitz rule g ha s continuous derivative
2π
g '(t ) =
∂
∫ ∂t ϕ (s, t ) ds
0
33
is well defined and
2π
=
∫
0

∂  f ( z + t (eis − z ))eis
− f ( z )  ds

is
∂t 
e −z

2π
=
∫
f '( z + t (eis − z ))eis ds
0
Let Φ( s ) = −i t −1 f ( z + t (eis − z )) then Φ '( s ) = f '( z + t (eis − z ))eis for 0 < t ≤ 1
g '(t ) = Φ (2π ) − Φ(0)
(By fundamental theorem of calculus for line integrals)
= −i t −1 f ( z + t (e 2π i − z )) + i t −1 f ( z + t (e0i − z ))
=0
g '(t ) = 0 for 0 < t ≤ 1
Since g is continuous on [0,1] we have g '(0) = 0
g '(t ) = 0 for 0 ≤ t ≤ 1
g is constant on [0,1]
g (1) = g (0)
2π
∫
0
2π
 f ( z + eis − z )eis

 f ( z )eis

− f ( z ) ds = ∫  is
− f ( z ) ds

is
e −z
e −z



0 
2π
 2π eis

= f ( z )  ∫ is
ds − ∫ ds 
0
0 e −z

= f ( z ) [ 2π − 2π ] = 0 .
Lemma 17 Let γ be a rectifiable curve in
and suppose that Fn and F are continuous
functions on {γ } . If F = u − lim Fn on {γ } then
∫γ F = lim ∫γ F
n
.
Proof. Since F = u − lim Fn , for given ∈> 0 there is an integer n0 > 0 such that
Fn ( w) − F ( w) < ∈
for all w ∈ {γ } and n ≥ n0 .
V (γ )
34
Therefore
∫γ F − ∫γ F
n
=
∫γ ( F − F )
n
≤ ∫ F ( w) − Fn ( w) dw
γ
≤∈
when n ≥ n0 .
The following theorem gives the Taylor’s series expansion of an analytic function:
∞
Theorem 18 Let
f be analytic in B(a, R ) . Then
f ( z ) = ∑ an ( z − a ) n for
z−a < R
n =0
where an =
1 (n)
f (a ) and this series has radius of convergence ≥ R .
n!
Proof. Since f is analytic in B(a, R ) , then there is 0 < r < R such that B(a, r ) ⊂ B(a, R) .
Let γ (t ) = a + reit , 0 ≤ t ≤ 2π then by Cauchy integral formula , we have
f ( z) =
1
f ( w)
dw for z − a < r .
∫
2π i γ w − z
…(1)
Now
1
1
1
=
=
.
w − z ( w − a) − ( z − a) w − a
1   z −a 
=
1−
w − a   w − a  
1
 z−a 
1− 
 w − a 
−1
−1
1   z −a 
1−
=
.
w − a   w − a  
n
∞
1
1
 z−a 
==
.
∑
w− z
( w − a ) n=0  w − a 
…(2)
35
Since {γ } is compact and f is continuous on {γ } , f bounded on
{γ }
Let M = sup { f ( w) : w ∈ {γ } }
Then
f ( w) ( z − a ) n
( z − a)n
f ( w)( z − a ) n
M
=
≤
( w − a )n +1
( w − a) ( w − a)n
( w − a ) n +1
M  ( z − a) 
≤ 

r  r 
n
n
M  ( z − a) 
Let M n =

 , then
r  r 
∞
∑M
n
< ∞ as
n =0
∞
By Weierstrass M- test the series
∑
n =0
z−a
<1
r
f ( w)( z − a)n
f ( w)
converges uniformly to
n +1
( w − a)
w− z
w ∈ {γ }
In view of (1), (2) and Lemma
f ( z) =
∞
1
f ( w)
1
f ( w)( z − a) n
=
dw
dw .
∑
2π i ∫γ w − z
2π i ∫γ n =0 ( w − a) n +1
1
f ( w)( z − a) n
∫ (w − a)n+1 dw
n = 0 2π i γ
∞
=∑
∞ 

1
f ( w)
dw
( z − a )n
= ∑

n +1
∫


n = 0  2π i γ ( w − a )

∞
= ∑ an ( z − a ) n where an =
n =0
1
f ( w)
dw , z − a < r
∫
2π i γ ( w − a ) n +1
Thus f has power series expansion in B(a, R )
an =
1 (n)
f (a ) , so that value of an is independent of γ , hence independent of r
n!
36
for
Moreover , as 0 < r < R is arbitrary , we have
∞
f ( z ) = ∑ an ( z − a ) n for z − a < R
n =0
clearly radius of convergence ≥ R .
∞
is analytic and a ∈ G then f ( z ) = ∑ an ( z − a ) n for z − a < R
Corollary 19 If f : G →
n =0
where R = d ( a, ∂G ) .
Proof. Since R = d ( a, ∂G ) = inf {d ( a, ∂G ) : z ∈ ∂G} = inf { z − a : z ∈ ∂G} , B (a, R ) ⊂ G
Since f is analytic in G , f is analytic on B (a, R )
Hence by Taylor’s theorem,
∞
f ( z ) = ∑ an ( z − a ) n
for
z−a < R
n =0
Corollary 20 If f : G →
is analytic then f is infinitely differentiable.
Proof. Suppose f : G →
is analytic, then for any a ∈ G , f has Taylor’s series expansion
about a
∞
f ( z ) = ∑ an ( z − a ) n
for
z − a < R , R = d ( a, ∂G )
n =0
Then by theorem, f is infinitely differentiable.
Corollary 21 If f : G →
is analytic and B (a, r ) ⊂ G then
f ( n ) (a) =
where γ (t ) = a + reit , 0 ≤ t ≤ 2π
Proof. Suppose f : G →
is analytic and B (a, r ) ⊂ G
Let γ (t ) = a + reit , 0 ≤ t ≤ 2π .
Then by Taylor’s theorem,
∞
f ( z ) = ∑ an ( z − a ) n where an =
n =0
37
1
f ( w)
dw
∫
2π i γ ( w − a ) n +1
n!
f ( w)
dw
∫
2π i γ ( w − a ) n +1
We also have,
an =
f ( n ) (a) =
1 (n)
f (a)
n!
n!
f ( w)
dw .
∫
2π i γ ( w − a ) n +1
Example 22 Evaluate the integral
eiz
it
∫γ z 2 dz where γ (t ) = re , 0 ≤ t ≤ 2π .
Solution: Let f ( z ) = eiz , then f is analytic function
We have
f (1) (0) =
1!
f ( z)
dz
∫
2π i γ ( z − 0)1+1
1! eiz
ie =
dz
2π i ∫γ z 2
i0
eiz
∫γ z 2 dz = −2π .
Example 28 Evaluate the integral
∫γ
sin z
dz where γ (t ) = reit , 0 ≤ t ≤ 2π .
3
z
Solution: Let f ( z ) = sin z , then f is analytic function.
We have
f (2) (0) =
2!
f ( z)
dz
∫
2π i γ ( z − 0) 2+1
− sin(0) =
Therefore
∫γ
1 sin z
dz
π i ∫γ z 3
sin z
dz = 0 .
z3
38
Example 23 Evaluate the integral
1
∫γ z − a dz
where γ (t ) = a + reit , 0 ≤ t ≤ 2π .
Solution: Let f ( z ) = 1 , then f is analytic function.
We have
0!
f ( z)
dz
∫
2π i γ ( z − a )0 +1
f (0) (a ) =
1=
1
1
dz
∫
2π i γ z − a
Therefore,
1
∫γ z − a dz = 2π i .
e z + sin z
Example 24 Evaluate the integral ∫
dz where γ (t ) = reit , 0 ≤ t ≤ 2π .
z
γ
Solution: Let f ( z ) = e z + sin z , then f is analytic function.
We have
f (0) (0) =
1!
f ( z)
dz
∫
2π i γ ( z − 0)0 +1
e0 + sin 0 =
Therefore
0! e z + sin z
dz
2π i ∫γ
z
e z + sin z
∫γ z dz = 2π i .
The following theorem is known as Cauchy’s Estimate
Theorem 25 Let f be analytic in B (a, R ) and suppose that
Then f ( n ) (a ) ≤
f ( z ) ≤ M for all f in B (a, R ) .
n !M
Rn
Proof. Since f is analytic in B (a, R ) , then there is 0 < r < R such that B (a, r ) ⊂ B (a, R ) .
Then by corollary,
f ( n ) (a) =
n!
f ( w)
dw where γ (t ) = a + reit , 0 ≤ t ≤ 2π .
n +1
∫
2π i γ ( w − a )
39
Therefore f ( n ) (a ) =
f ( n ) (a) ≤
n!
f ( w)
dw
∫
2π i γ ( w − a ) n +1
f ( w)
≤
n!
2π
≤
n! M
dw
2π r n +1 ∫γ
≤
n! M
2π r .
2π r n +1
∫γ (w − a)
n +1
dw
n !M
.
rn
Since 0 < r < R is orbitrary , letting r → R − we get
f ( n ) (a) ≤
n !M
.
Rn
Cauchy’s estimate leads us to Cauchy’s Theorem:
Theorem 26 Let f be analytic in the disk B (a, R ) and suppose that γ is a closed rectifiable
curve in B (a, R ) . Then
∫γ f
= 0.
Proof. Since f is analytic in B (a, R ) , it has power series expansion
∞
f ( z ) = ∑ an ( z − a ) n
for z − a < R .
…(1)
n =0
Let
∞
∞
 a 
 a 
F ( z ) = ∑  n  ( z − a )n +1 = ( z − a )∑  n  ( z − a )n
n=0  n + 1 
n=0  n + 1 
Since lim ( n + 1)
a
lim Sup n
n +1
1
n
1
n
=1
= lim Sup an
1
n
 1 
lim 

 n +1 
1
n
= lim Sup an
Thus series (1) and (2) have same radius of convergence.
Therefore, F is defined and analytic in B (a, R ) .
40
1
n
.
…(2)
Hence
∞
∞
 a 
F '( z ) = ∑  n  ( n + 1) ( z − a ) n = ∑ an ( z − a )n = f ( z ) .
n=0  n + 1 
n =0
i.e. F is primitive of f .
If γ :[a, b] →
then
∫γ f =∫γ F ' = F (γ (b)) − F (γ (a)) = 0 , Since γ
is a closed curve γ (a ) = γ (b) .
EXERCISES
1) Evaluate the integral
2) Evaluate
the
1
∫γ z dz , where γ (t ) = e
integral
∫γ z dz, ,
n
int
where
, 0 ≤ t ≤ 2π , n is some positive integer.
γ
is
a
closed
polygonal
curve
[1-i, 1+i, -1+i, -1-i, 1-i].
3) Let γ , δ be polygons [1, 1+i, i] and [1, i] respectively. Evaluate the integral ∫ z 2 dz
over γ as well as δ .
4) Evaluate
eiz
it
∫γ z 2 dz , where γ (t ) = e , 0 ≤ t ≤ 2π .
5) Evaluate
∫γ
sin( z )
dz , where γ (t ) = eit , 0 ≤ t ≤ 2π .
3
z
6) Let G be connected set and f : G →
be analytic function. If f(z) is real for all z in
G, the prove that f is constant function.
7) Prove above exercise for a) f(z) is imaginary number for all z and b) f(z) with
constant modulus.
41
UNIT - IV
FUNDAMENTAL THEOREM OF ALGEBRA AND
MAXIMUM MODULUS THEOREM
In this unit we prove Liouville’s theorem use it to prove fundamental theorem of
algebra. We also prove maximum modulus theorem.
Definition 1 An entire (integral) function is a function which is defined and analytic in the
whole complex plane .
Note 1. f ( z ) = e z ,sin z, cos z are entire functions.
2. All polynomials are entire functions.
∞
Theorem 2 If f
is entire function then f has power series expansion f ( z ) = ∑ an z n with
n =0
infinite radius of convergence.
Proof. For any R > 0 , B (a, R ) ⊂
.Then f is analytic in B (a, R ) .
By Taylor’s theorem,
∞
f ( z ) = ∑ an z n , for z < R .
n =0
Since R > 0 is arbitrary, radius of convergence is infinite.
Following theorem is known as Liouville’s theorem
Theorem 3 If f is bounded and entire function then f is constant.
Proof. Since f is bounded and entire function,
R > 0 , f is analytic in B (a, R ) .
By Cauchy’s Estimate, we have f ( n ) (a ) ≤
n !M
Rn
42
f ( z) ≤ M
∀ z∈
and for a ∈
,
In particular for n = 1 we have
f ' (a) ≤
n !M
R
Since R is arbitrary , as R → ∞ , we get f ' (a ) ≤ 0 .
Therefore, f ' (a ) = 0 for any a ∈ .
Thus f is constant.
Thus we can prove the Fundamental theorem of Algebra:
Theorem 3 If p ( z ) is a non constant polynomial then there is a complex number a with
p(a) = 0 .
Proof. Let p( z ) is a non constant polynomial and that p( z ) ≠ 0 for any z ∈
Let f ( z ) =
1
then f is entire function .
p( z )
.
( p( z ) is entire and p( z ) ≠ 0 )
Since p( z ) is non constant polynomial, assume that p ( z ) = z n + a1 z n −1 + a2 z n − 2 + ... + + an
lim p ( z ) = lim z n (1 + a1 z −1 + a2 z −2 + ... + an z − n ) = ∞
z →∞
z →∞
lim f ( z ) = lim
z →∞
z →∞
1
=0
p( z )
Therefore, for ∈= 1 there is R > 0 such that
f ( z ) − 0 < 1 , whenever z > R .
That is f ( z ) < 1 , whenever z > R .
Since f is continuous on closed bounded disk B (0, R ) ⊂
, f is bounded on B (0, R ) .
Therefore, there is M > 0 such that f ( z ) ≤ M for z ∈ B (0, R )
That is f ( z ) ≤ M , whenever z ≤ R .
Thus f ( z ) ≤ max {1, M } , for all z ∈
.
This f is bounded entire function.
Hence by Liouville’s theorem f is constant and consequently p is constant.
Which contradicts our assumption.
Hence the theorem.
43
Definition 4 Let f : G →
be analytic and a ∈ G satisfies f (a ) = 0 then a is zero of f of
multiplicity m ≥ 1 if there is an analytic function g : G →
such that f ( z ) = ( z − a ) m g ( z ) ,
where g (a ) ≠ 0 .
Corollary 5 If
p ( z ) is a polynomial and a1 , a2 ,..., am are its
zeros with a j having
multiplicity k j then p ( z ) = c( z − a1 )k1 ...( z − am ) km for some constant c and k1 + k2 + ... + km is
the degree of p .
Proof. Since a1 , a2 ,..., am are zeros of p ( z ) having multiplicities k1 , k2 ,..., km respectively,
there exists a polynomial
g ( z ) such that
p ( z ) = ( z − a1 ) k1 ...( z − am ) km g ( z ) ,
where
g (a j ) ≠ 0, ( 1 ≤ j ≤ m ) .
Therefore by fundamental theorem of algebra, g ( z ) is forced to be constant.
Let g ( z ) = c for some c ∈
p ( z ) = c ( z − a1 ) k1 ...( z − am )km .
Obviously, degree of p ( z ) is k1 + k2 + ... + km .
Theorem 6 Let G be a connected open set and let
f :G →
the following are equivalent statements:
(a) f ≡ 0 ;
(b)
{ z ∈ G : f ( z) = 0 }
has limit point in G ;
(c) there is a point a in G such that f n (a ) = 0 for each n ≥ 0 .
Proof.
(a ) ⇒ (b)
suppose f ≡ 0 , then
{ z ∈ G : f ( z ) = 0 } = G , which is open.
Hence, every point of G is a limit point of G .
Thus
{ z ∈ G : f ( z) = 0 }
has limit point in G .
(b) ⇒ (c)
44
be analytic function. Then
Suppose that, A = { z ∈ G : f ( z ) = 0 } has limit point a in G , then there is a sequence { zn }
of points in A such that a = lim zn .
( zn ∈ A ∴ f ( z n ) = 0 )
Since f is continuous , f (a ) = lim f ( zn ) = 0
Now suppose that, there is an integer n ≥ 1 such that f (a ) = f '(a ) = ... = f ( n −1) (a ) = 0
and f ( n ) (a ) ≠ 0 .
Since f is analytic in G , there is R > 0 such that f is analytic in B ( a; R ) ⊂ G .
By Taylor’s theorem
∞
f ( z ) = ∑ ak ( z − a ) k
for z − a < R
where
k =0
ak =
1 (k )
f (a) .
k!
Since f (a ) = f '(a ) = ... = f ( n −1) (a ) = 0 and f ( n ) (a ) ≠ 0 we have
∞
∞
k =n
k =n
f ( z ) = ∑ ak ( z − a )k = ( z − a )n ∑ ak ( z − a ) k − n
∞
Let g ( z ) = ∑ ak ( z − a ) k − n , then g is analytic in B ( a; R ) and
k =n
f ( z ) = ( z − a ) n g ( z ) and g (a ) = an ≠ 0 .
Since g is continuous in B ( a; R ) , there is R > r > 0 such that g ( z ) ≠ 0 in B ( a; r ) .
As a is limit point of A , B ( a; r ) ∩ A − {a} ≠ φ .
Let b ∈ B ( a; r ) ∩ A − {a} , then b ∈ B ( a; r ) and b ∈ A − {a} ⊂ A .
Therefore, b ∈ B ( a; r ) ⇒
g (b) ≠ 0 and b ∈ A − {a} ⊂ A ⇒
which gives the contradiction to our assumption .
Hence no such integer n ≥ 1 can be found .
Thus f n (a ) = 0 for each n ≥ 0 .
(b) ⇒ (c)
Suppose there is a in G such that f n (a ) = 0 for each n ≥ 0 .
Let H = { z ∈ G : f ( n ) ( z ) = 0 } then H ≠ φ .
45
f (b) = 0 ⇒ g (b) = 0
Now we claim that H is both open and closed :
B ( a; R ) ⊂ G R > 0 . Then f is analytic in
Let a ∈ H , then there is R > 0 such that
B ( a; R ) ⊂ G .
By Taylor’s theorem
∞
f ( z ) = ∑ an ( z − a ) n
for z − a < R
where
n =0
an =
1 (n)
f (a) .
n!
Since f n (a ) = 0 for each n ≥ 0 , each an = 0 .
f ( z ) = 0 in B ( a; R ) i.e. f ≡ 0 in B ( a; R ) .
f n ( z ) = 0 in B ( a; R )
B ( a; R ) ⊆ H
Thus for a ∈ H , then there is R > 0 such that B ( a; R ) ⊆ H .
Hence H is open.
Now let z be limit point of H , then there is a sequence { zm } in H such that z = lim zm
Since f ( n ) is continuous , f ( n ) ( z ) = lim f ( n ) ( zn ) = 0
Therefore f ( n ) ( z ) = 0 implies z ∈ H .
Thus H ⊆ H .
Hence H is closed.
Thus H is open as well as closed subset of connected set G .
Hence by property of connectedness H = G .
Therefore, f ( n ) ( z ) = 0 ∀ z ∈ G , n ≥ 0
That is f ( z ) = 0 ∀ z ∈ G
Hence f ≡ 0 on G .
Corollary 7 If f and g are analytic on a region G , then f ≡ g iff
has a limit point in G .
Proof. Let h( z ) = f ( z ) − g ( z ) ∀ z ∈ G , which is analytic in G .
46
{ z ∈ G : f ( z) = g ( z) }
Then h ≡ 0 on G ⇔
{ z ∈ G : h( z ) = 0 }
i.e. f ( z ) − g ( z ) = 0 on G ⇔
i.e. f ( z ) = g ( z ) on G ⇔
i.e. f ≡ g on G ⇔
has a limit point in G .
{ z ∈ G : f ( z) − g ( z) = 0 }
{ z ∈ G : f ( z) = g ( z) }
{ z ∈ G : f ( z) = g ( z) }
has a limit point in G .
has a limit point in G .
has a limit point in G .
Corollary 8 If f is analytic on an open connected set G and f is not identically zero then
for each a in G with f (a ) = 0 there is an integer n ≥ 1 and an analytic function g : G →
such that g (a ) ≠ 0 and f ( z ) = ( z − a ) n g ( z ) for all z in G . That is each zero of f has finite
multiplicity.
Proof. Let f be analytic on an open connected set G . Since f ≠ 0 and f (a ) = 0
for some a in G , there is positive integer n ≥ 1 such that f (a ) = f '(a ) = ... = f ( n −1) (a ) = 0
and f ( n ) (a ) ≠ 0 .
Now we define g : G →
, by
f ( z)
( z − a)n
for
z≠a
f ( n ) (a)
=
n!
for
z=a
g ( z) =
…(1)
Therefore g is analytic on G − {a} .
Now to show that g is analytic on G it need only to show g is analytic in a neighborhood
of a .
Since f is analytic in G , there is R > 0 such that f is analytic in B ( a; R ) ⊂ G .
By Taylor’s theorem
∞
f ( z ) = ∑ ak ( z − a ) k
for z − a < R
where
k =0
ak =
1 (k )
f (a) .
k!
Since f (a ) = f '(a ) = ... = f ( n −1) (a ) = 0 and f ( n ) (a ) ≠ 0 we have
∞
∞
k =n
k =n
f ( z ) = ∑ ak ( z − a )k = ( z − a )n ∑ ak ( z − a ) k − n
47
∞
Let h( z ) = ∑ ak ( z − a ) k − n , then h is analytic in B ( a; R ) and
k =n
f ( z ) = ( z − a ) n h( z ) and h(a ) = an =
f ( n ) (a)
.
n!
Thus from (1) g ≡ h in B ( a; R )
Therefore g is analytic in B ( a; R ) .
Hence g is analytic in G with f ( z ) = ( z − a ) n g ( z ) and g (a ) = h(a ) = an ≠ 0 .
Corollary 9 If f : G →
is analytic and not constant, a ∈ G and f (a ) = 0 then there is an
R > 0 such that B ( a; R ) ⊂ G and f ( z ) ≠ 0 for 0 < z − a < R . That is zeros of are isolated.
Proof. Let f : G →
be non-constant analytic function with f (a ) = 0 for some a ∈ G .
Then by corollary there is an analytic function g : G →
and an integer n ≥ 1 such that
f ( z ) = ( z − a ) n g ( z ) and g (a ) ≠ 0 .
Since g is analytic, g is continuous on G .
Therefore, there is R > 0 such that g ( z ) ≠ 0 in B ( a; R ) ⊂ G i.e for z − a < R
Hence f ( z ) = ( z − a ) n g ( z ) ≠ 0 for 0 < z − a < R .
We now prove Maximum Modulus Theorem:
Theorem 10 Let G is a region and f : G →
is an analytic function such that there is a
point a in G with f (a) ≥ f ( z ) for all z in G , then f is constant.
is analytic function, there is r > 0 such that B ( a; r ) ⊂ G .
Proof. Since f : G →
Then by Cauchy Integral formula
f (a) =
=
=
1
f ( w)
dw for z − a < r and γ (t ) = a + reit , 0 ≤ t ≤ 2π .
∫
2π i γ w − a
1
2π i
1
2π
2π
∫
0
f (a + reit ) it
rie dt
a + reit − a
2π
∫
f (a + reit ) dt
0
48
1
f (a) ≤
2π
1
≤
2π
2π
∫
f (a + reit ) dt
0
2π
∫
f (a ) dt
as
f (a) ≥ f ( z ) for all z in G .
0
= f (a )
Therefore
1
2π
f (a) ≤
f (a) =
1
2π
1
2π
2π
∫
f (a + reit ) dt ≤ f (a )
0
2π
∫
f (a + reit ) dt
0
2π
∫  f (a) −
f (a + reit )  dt = 0
…(1)
0
Since f (a ) − f (a + reit ) ≥ 0 for all t
Therefore from (1) we have f (a ) = f (a + reit ) for all t .
Let f (a ) = α , then
f (a + reit ) = α for all t .
Since r > 0 is arbitrary, we have f ( z ) = α for all z in B ( a; r ) .
That is f maps whole disk B ( a; r ) ⊂ G into the circle z = α where f (a ) = α .
Therefore f has constant modulus on B ( a; r ) and hence f is constant on B ( a; r ) .
Let f ( z ) = c on B ( a; r ) ⊂ G then
{ z ∈ G : f ( z ) = c } ⊇ B ( a; r )
has a limit point in G .
Thus f ( z ) = c on G , i.e. f is constant on G .
Theorem 11 If γ : [ 0,1] →
is closed rectifiable curve a ∉ {γ } , then
1
1
dz is an integer.
∫
2π i γ z − a
Proof. Define g : [ 0,1] →
t
g (t ) = ∫
0
by
γ '( s )
γ ( s) − a
ds
49
where γ is closed rectifiable curve so that g is well defined .
γ '( s)
1
Therefore g (0) = 0 and g (1) = ∫
0
Also g '(t ) =
γ '(t )
γ (t ) − a
γ ( s) − a
ds = ∫
γ
1
dz .
z−a
for 0 ≤ t ≤ 1 .
d − g (t )
e
(γ (t ) − a )  = e− g (t )γ '(t ) − e − g (t ) g '(t ) (γ (t ) − a)
dt
Then


γ '(t )
= e− g (t ) γ '(t ) −
(γ (t ) − a ) 
γ (t ) − a


=0
Therefore e− g (t ) (γ (t ) − a ) is constant function.
e− g (0) (γ (0) − a ) = e − g (1) (γ (1) − a )
Hence
e0 = e g (1)
( γ (0) = γ (1) )
e2π ik = e g (1)
( k is integer )
Therefore
g (1) = 2π ik
Therefore,
2π ik = ∫
γ
Hence
1
dz
z−a
1
1
dz = k .
∫
2π i γ z − a
Definition 12 If γ is a closed rectifiable curve in
then for a ∉ {γ } . Then the integer
1
1
dz is called the index of γ with respect to the point a .
∫
2π i γ z − a
Definition 13 A subset D of a metric space X is called component of X , if it is maximal
connected subset of X.
50
Theorem 14 Let γ be a closed rectifiable curve in
. Then
a) n(γ ; a ) is constant for a belonging to a component of G =
− {γ } .
b) n(γ ; a ) = 0 for a belonging to the unbounded component of G .
Proof. Define f : G →
f (a) =
by
1
1
dz = n(γ , a )
∫
2π i γ z − a
∀ in G .
Claim: f is continuous.
Let a ∈ G and r = d ( a, {γ } ) > 0 .
For any ∈> 0 we choose δ > 0 such that b − a < δ < r .
2
Therefore
f (b) − f (a ) =
1  1
1 
−
dz
∫

2π i γ  z − b z − a 
b−a
≤
1
2π
≤
b−a
2π
∫γ ( z − b )( z − a ) dz
Now z − a ≥ r > r
dz
∫γ ( z − b ) ( z − a )
2
and z − b = ( z − a ) − (b − a ) ≥ z − a − b − a > r − r = r .
2
2
Thus
f (b) − f (a ) <
δ 2 2
2δ
. dz = 2 V (γ ) .
∫
2π γ r r
πr
 r ∈π r 2 
Therefore, for any ∈> 0 there is 0 < δ < min  ,
,
 2 2 V (γ ) 
whenever b − a < δ .
Thus f is continuous .
51
such that
f (b) − f (a) <∈ ,
a) Let D be component of G then D is open and connected .
Since f is continuous, f ( D) is connected.
Since f is integer valued and subset of set of all integers which are connected are precisely
singleton sets .
Therefore f ( D) = {k } for some integer k .
That is f (a ) = k
∀ in D .
Hence n(γ ; a ) is constant for a belonging to D .
b) Let U be unbounded component of G =
− {γ } , then there is R > 0 such that
{ z ∈G : z > R } ⊆ U .
For ∈> 0 choose a such that a > R and z − a >
n(γ ; a ) =
≤
V (γ )
for all z on {γ } then
2π ∈
1
1
dz
∫
2π i γ z − a
1
2π
∫γ
1
1 2π ∈
dz ≤
dz =∈
z−a
2π V (γ ) ∫γ
Therefore n(γ ; a) <∈
Since n(γ ; a ) is an integer n(γ ; a) <∈ ⇒ n(γ ; a ) = 0 for some a in U .
Since f (a ) = n(γ ; a ) is constant on U , we must have n(γ ; a ) = 0 for all a in U .
52
EXERCISES
1) Let f : C  C be entire function. Suppose for some R > 0,z > R
implies
 f(z) ≤ Mzn, for some constant M. Then prove that f is a polynomial of
degree at most n.
2) Let f : G  C be analytic function defined on a region G with f(a) ≤ f(z), for all z
in G. Show that either f ≡ 0 or f is constant function.
3) Let f and g be nalytic functions defined on the region G. If f.g = 0 on G, prove that
either f ≡ 0 or g ≡ 0.
4) Show by an example that n(γ ; a ) = k for a closed rectifiable curve γ in C, where
a ∉ {γ }.
53
UNIT - V
WINDING NUMBERS AND CAUCHYS
INTEGRALTHEOREM
1
1
dz is an integer. We shall denote this integer by
∫
2π i γ z − a
In the last unit we prove that
n(γ ; a ) and called it is a winding number or Index of a closed curve γ around a. In this unit
we discuss Cauchy’s integral formulae.
Lemma 1 Let γ be rectifiable curve and suppose φ is a function defined and continuous on
{γ } . For each
m ≥ 1 let
Fm ( z ) = ∫
γ
φ ( w)
(w − z)
m
dw for z ∉ {γ } . Then each Fm is analytic on
− {γ } and Fm' ( z ) = mFm +1 ( z ) .
Proof. Let Fm ( z ) = ∫
φ ( w)
γ (w − z)
m
dw for z ∉ {γ } where φ is continuous function and γ is
rectifiable curve .
First we claim that each Fm is continuous:
Let a ∈ G =
− {γ } and r = d ( a, {γ } ) > 0 .
For any ∈> 0 we choose δ > 0 such that z − a < δ < r .
2
Therefore
 φ ( w)
φ ( w) 
Fm ( z ) − Fm (a ) = ∫ 
dw
−
m
m 
−
−
w
z
w
a
(
)
(
)

γ 


≤ ∫ φ ( w)
γ
1
(w − z)
m
−
1
(w − a)
m
dw
…(1)
Since φ is continuous function on compact set {γ } , we have M = sup { φ ( w) : w ∈ {γ }} .
54
And
1
(w − z)
m
−
 1
1  m −1
1
=
−
∑
m − k −1
k
(w − a)
 ( w − z ) ( w − a )  k =0 ( w − z )
1
(w − a)
m
m −1
= ( z − a )∑
k =0
Also w − a ≥ r > r
Therefore
2
1
(w − z)
m
−
1
(w − z)
m−k
(w − a)
k +1
and w − z = ( w − a ) − ( z − a ) ≥ w − a − z − a > r − r = r .
2
2
m −1
1
(w − a)
m
= ( z − a )∑
k =0
1
(w − z)
(w − a)
m−k
m −1
≤ ( z − a) ∑
k =0
<δ
m −1
∑
k =0
1
(w − z)
m−k
1
( 2) ( 2)
r
k +1
m−k
r
(w − a)
k +1
( r)
m +1
=δ 2
k +1
m
Thus (1) gives
( r)
Fm ( z ) − Fm (a) < ∫ M δ 2
γ
m +1
( r)
m dw = mM δ 2
m +1
V (γ ) .
( )
m +1


r
r ∈ 2

, such that Fm ( z ) − Fm (a) <∈ ,
Therefore, for any ∈> 0 there is 0 < δ < min  ,
2 mM V (γ ) 




whenever z − a < δ .
Thus Fm is continuous on G =
− {γ } for any m ≥ 1 .
Now to show Fm' ( z ) = mFm +1 ( z ) :
Consider
 φ ( w)
φ ( w) 
Fm ( z ) − Fm (a ) = ∫ 
−
dw
m
m 
w
−
z
w
−
a
(
)
(
)

γ 


55
φ ( w)
m +1
= ( z − a)∫ ∑
γ k =0
(w − z)
m−k
(w − a)
k +1
dw
Thus
Fm ( z ) − Fm (a ) m +1
=∑
( z − a)
k =0
φ ( w) ( w − a )k +1 


∫γ ( w − z )m−k dw .
Since a ∉ {γ } and φ ( w) ( w − a )
k +1
{γ }
continuous on
for each k , each integral
φ ( w) ( w − a )k +1 


∫γ ( w − z )m−k dw is continuous .
Hence letting z → a we get
m −1
F ( z ) − Fm (a )
lim m
= lim ∑
z →a
z →a
( z − a)
k =0
φ ( w) ( w − a )k +1 


∫γ ( w − z )m−k dw
φ ( w) ( w − a )k +1 
m −1
 dw =
= ∑ lim ∫ 
∑
m−k
z →a
k =0
k =0
(w − z)
γ
m −1
lim
z →a
Fm ( z ) − Fm (a) m −1
=∑
( z − a)
k =0
φ ( w)
∫γ ( w − a )
m +1
dw = m ∫
γ
φ ( w)
(w − a)
Therefore,
Fm' (a ) = mFm +1 (a ) for all a ∈
− {γ } .
Thus, Fm is differentiable for any m ≥ 1 .
Since Fm +1 is continuous, F 'm is continuous .
Therefore Fm is continuously differentiable.
Hence Fm is analytic on
− {γ } .
56
m +1
φ ( w) ( w − a )k +1 


∫γ ( w − a )m−k dw
dw
We now prove Cauchy‘ s Integral formulae:
Theorem 2 ( First Version) Let G be an open subset of the plane and
f :G →
be an
analytic function. If γ is a closed rectifiable curve in G such that n(γ ; w) = 0 for all
− G , then for a in G − {γ }
w in
n(γ ; a ) f (a ) =
Proof. Define φ : G × G →
1
f ( z)
dz
∫
2π i γ z − a
by
f ( w) − f ( z )
w− z
φ ( z, w) =
if z ≠ w
if z = w .
= f '( z )
Then φ is continuous .
Let H = {w ∈ : n(γ ; w) = 0} . Since n(γ ; w) is a continuous integer valued function , H is
open. Moreover
− G ⊆ H . Thus
Now, define g :
→
=G∪H .
by
g ( z ) = ∫ φ ( z , w) dw if z ∈ G
γ
=∫
γ
f ( w)
dw
w− z
if z ∈ H .
If z ∈ G ∩ H then
∫γ φ ( z, w) dw = ∫γ
f ( w) − f ( z )
dw
w− z
=∫
f ( w)
1
dw − f ( z ) ∫
dw
w− z
γ w− z
=∫
f ( w)
dw − f ( z ) n(γ ; z ) 2π i
w− z
=∫
f ( w)
dw − f ( z ). 0 .2π i
w− z
=∫
f ( w)
dw .
w− z
γ
γ
γ
γ
Therefore g is well defined function.
Thus by lemma g is analytic on
, hence g is entire function.
57
1
= 0 uniformly for w ∈ {γ } .
z →∞ w − z
Since H contains neighborhood of infinity, we have lim
Since {γ } is compact, f is bounded on {γ } .
Hence there is M > 0 such that
Therefore,
f ( w)
f ( w)
∫γ w − z dw ≤ ∫γ w − z
≤M∫
γ
Hence lim g ( z ) = lim ∫
z →∞
z →∞
γ
f ( w) ≤ M for all w ∈ {γ } .
dw
1
dw
w− z
f ( w)
dw = 0
w− z
Therefore, there is R > 0 such that g ( z ) ≤ 1 for z > R .
Since, g continuous on compact set B (0, R ) , g is bounded on B (0, R ) .
Thus g is a bounded entire function. Hence by Liouville’s Theorem g is constant.
Since lim g ( z ) = 0 we must have g ≡ 0 .
z →∞
Thus
∫γ
f ( w) − f (a )
dw = 0 for all a in G − {γ } .
w−a
f ( w)
1
∫γ w − a dw = f (a)∫γ w − a dw
f ( w)
∫γ w − a dw = f (a).2π i.n(γ ; a)
for all a in G − {γ } .
for all a in G − {γ } .
Thus
n(γ ; a ) f (a ) =
1
f ( z)
dz
∫
2π i γ z − a
for all a in G − {γ } .
58
Theorem 3 (Second Version) Let G be an open subset of the plane and f : G →
analytic function.
If
γ 1 , γ 2 ,..., γ m
are closed rectifiable curves in
n(γ 1 ; w) + n(γ 2 ; w) + ... + n(γ m ; w) = 0 for all w in
m
be an
G such that
− G , then for a in G − {γ }
m
1
f ( z)
dz
∫
k =1 2π i γ k z − a
f (a )∑ n(γ k ; a ) = ∑
k =1
Proof. Define φ : G × G →
φ ( z , w) =
f ( w) − f ( z )
w− z
by
if z ≠ w
if z = w .
= f '( z )
Then φ is continuous .
Let H = {w ∈ : n(γ 1 ; w) + n(γ 2 ; w) + ... + n(γ m ; w) = 0} . Since n(γ ; w) is a continuous integer
valued function , H is open. Moreover
→
Now, define g :
− G ⊆ H . Thus
by
m
g ( z ) = ∑ ∫ φ ( z , w) dw if z ∈ G
k =1 γ k
m
=∑∫
k =1 γ k
f ( w)
dw
w− z
if z ∈ H .
If z ∈ G ∩ H then
m
m
∑ ∫ φ ( z, w) dw = ∑ ∫
k =1 γ k
k =1 γ k
f ( w) − f ( z )
dw
w− z
m 

f ( w)
1
= ∑∫
dw − f ( z ) ∫
dw
k =1 

γk w − z
γ k w − z
m 

f ( w)
= ∑∫
dw − f ( z ) n(γ k ; z ) 2π i 
k =1 

γ k w − z
59
=G∪H .
m
=∑ ∫
k =1 γ
k
m
=∑ ∫
k =1 γ
k
m
m
f ( w)
f ( w)
dw − f ( z )2π i ∑ n(γ k ; z ) = ∑ ∫
dw − f ( z )2π i.0
w− z
k =1
k =1 γ k w − z
f ( w)
dw .
w− z
Therefore g is well defined function.
Thus by lemma g is analytic on
, hence g is entire function.
Since H contains neighborhood of infinity, we have lim
z →∞
1
= 0 uniformly for w ∈ {γ k } .
w− z
Since each {γ k } is compact , f is bounded on {γ k } .
Hence lim g ( z ) = 0
z →∞
Therefore, there is R > 0 such that g ( z ) ≤ 1 for z > R .
Since g continuous on compact set B (0, R ) , g is bounded on B (0, R ) .
Thus g is a bounded entire function. Hence by Liouville’s Theorem g is constant.
Since lim g ( z ) = 0 we must have g ≡ 0 .
z →∞
m
Thus
∑∫
k =1 γ k
m
∑∫
k =1 γ k
m
∑∫
k =1 γ m
m
f ( w) − f (a )
dw = 0 for all a in G − ∪ {γ k } .
k =1
w−a
m
m
f ( w)
1
dw = f (a )∑ ∫
dw for all a in G − ∪ {γ k } .
k =1
w−a
k =1 γ k w − a
m
m
f ( w)
dw = f (a ).2π i ∑ n(γ k ; a ) for all a in G − ∪ {γ k } .
k =1
w−a
k =1
Thus
m
m
1
f ( z)
dz
∫
k =1 2π i γ k z − a
f (a )∑ n(γ k ; a ) = ∑
k =1
60
m
for all a in G − ∪ {γ k } .
k =1
Theorem 4 Let G be an open subset of the plane and f : G →
be an analytic function. If
γ 1 , γ 2 ,..., γ m are closed rectifiable curves in G such that n(γ 1; w) + n(γ 2 ; w) + ... + n(γ m ; w) = 0
− G , then
for all w in
m
∑∫ f
=0
k =1 γ k
m
− ∪ {γ k } and
Proof. Let a ∈
F ( z ) = ( z − a) f ( z ) .
k =1
Then F is analytic in G .
Hence by Cauchy’s integral theorem,
m
m
1
F ( z)
dz
∫
k =1 2π i γ k z − a
F (a )∑ n(γ k ; a ) = ∑
k =1
m
m
2π i.0.∑ n(γ k ; a ) = ∑ ∫
k =1
k =1 γ k
( z − a) f ( z )
dz
z−a
m
∑ ∫ f ( z )dz = 0
k =1 γ k
m
Thus
∑∫ f
=0 .
k =1 γ k
Theorem 5 Let G be an open subset of the plane and f : G →
be an analytic function. If
γ 1 , γ 2 ,..., γ m are closed rectifiable curves in G such that n(γ 1; w) + n(γ 2 ; w) + ... + n(γ m ; w) = 0
for all w in
− G , then for a in G − {γ }
m
m
1
f ( z)
dz
∫
k =1 2π i γ k z − a
f ( n ) (a )∑ n(γ k ; a ) = n !∑
k =1
Proof. By Cauchy Integral formula we have
m
m
1
f ( z)
dz
∫
k =1 2π i γ k z − a
f (a )∑ n(γ k ; a ) = ∑
k =1
61
m
for all a in G − ∪ {γ k } .
k =1
Hence
dn
n
k =1 da
m
m
f ( n ) (a )∑ n(γ k ; a ) = ∑
k =1
 1
f ( z) 
dz 

∫
2
i
z
−
a
π
γk


1
∂ n  f ( z) 
 dz
∫ n
k =1 2π i γ k ∂a  z − a 
m
=∑
m
n!
f ( z)
dz
n +1
∫
k =1 2π i γ k ( z − a )
=∑
Therefore,
m
m
1
f ( z)
dz .
∫
k =1 2π i γ k z − a
f (a )∑ n(γ k ; a ) = ∑
k =1
Corollary 6 Let G be an open subset of the plane and f : G →
be an analytic function. If
γ is a closed rectifiable curve in G such that n(γ ; w) = 0 for all w in
− G , then for a in
G − {γ }
f ( n ) (a ) n(γ ; a ) =
n!
f ( z)
dz .
∫
2π i γ ( z − a )n +1
Definition 7 A closed polygonal path having three sides is called triangular path.
Theorem 8 Morera’s Theorem Let G be a region and let f : G →
be a continuous
function such that ∫ f = 0 for every triangular path T in G ; then f is analytic in G .
T
Proof. To prove that f is analytic in G we have to prove that f is analytic on each open
disk contained in G . Hence without loss of generality we may assume that G = B (a; R ) .
Now define F : G →
F ( z) =
∫
by
f ( w)dw
where [ a, z ] is the line segment joining a to z .
[ a. z ]
Fix z0 ∈ B (a; R ) , then for any z in G , T = [ a, z , z0 , a ] be a triangular path in G .
62
Therefore by hypothesis
∫f
=0 ⇒
∫
f+
[a, z ]
T
⇒
∫
∫
f+
[ z , z0 ]
f =
[a, z ]
⇒ F ( z) =
∫
f =0
[ z0 , a ]
∫
f+
∫
f ( w)dw + F ( z0 )
[ z0 , z ]
∫
f
[ a , z0 ]
[ z0 , z ]
Thus
F ( z ) − F ( z0 )
1
− f ( z0 ) =
f ( w)dw − f ( z0 )
z − z0
z − z0 [ z0∫, z ]
=
1
z − z0 [ z0∫, z ]
[ f ( w) − f ( z0 )] dw
Since f is continuous in G , for any ∈> 0 there is δ > 0 , such that
f ( w) − f ( z0 ) <∈ whenever w − z0 < δ .
Therefore
F ( z ) − F ( z0 )
1
− f ( z0 ) ≤
z − z0
z − z0
∫
f ( w) − f ( z0 ) dw <
[ z0 , z ]
∈
z − z0
∫
[ z0 , z ]
Thus
lim
z → z0
F ( z ) − F ( z0 )
= f ( z0 )
z − z0
F '( z0 ) = f ( z0 )
Since z0 is arbitrary , we have F ' ≡ f in G .
Since f is continuous, F ' is continuous on G
Therefore F is continuously differentiable, that is F is analytic.
Hence F ' ≡ f is also analytic in G .
63
dw =∈
Singularities
Definition 9 A function f has an singularity at z = a if f is not analytic at z = a .
1 sin z 1z
, e has an singularity at z = 0 .
Ex. ,
z z
Ex.
1
has an singularity at the points z =
( z)
cos 1
2
(2n + 1)π
, n = 0, ±1, ±2,...
Definition 10 A function f has an isolated singularity at z = a if there is R > 0 such that f
is analytic in B ( a; R ) − {a} ,otherwise z = a is non-isolated singularity of f .
Ex.
Ex.
1
, z=
( z)
sin π
1
are isolated singularities and z = 0 is non-isolated singularity.
n
1
, z = 1, 2 are isolated singularities.
( z − 1)( z − 2)
There are three kinds of isolated singularities
A) Removable singularity
B) Pole
C) Essential Singularity
Definition 11 An isolated singularity at z = a of a function f is removable singularity if
there is R > 0 and an analytic function g : B ( a, R ) →
such that g ( z ) = f ( z ) in
0< z−a < R.
Ex. f ( z ) =
sin z
has removable singularity at z = 0 .
z
Ex. f ( z ) =
z
has removable singularity at z = 0 .
e −1
z
Theorem 12 If f has an isolated singularity at z = a , then the point z = a is removable
singularity iff lim( z − a) f ( z ) = 0 .
z →a
64
Proof. Suppose z = a is removable singularity, then there is R > 0 and an analytic function
g : B ( a, R ) →
such that g ( z ) = f ( z ) in 0 < z − a < R .
Therefore, lim( z − a) f ( z ) = lim( z − a) g ( z ) = 0.g (a) = 0 .
z →a
( since g is continuous )
z →a
Conversely suppose that lim( z − a) f ( z ) = 0 . Since f has an isolated singularity at z = a ,
z →a
there is R > 0 such that f is analytic in B ( a; R ) − {a} .
 ( z − a ) f ( z ) if z ≠ a
Define h( z ) = 
if z = a
0
Clearly h is analytic in B ( a; R ) − {a} and continuous at z = a .
Now to prove f has removable singularity at z = a we have to prove that h is analytic in
B ( a; R ) .
Claim: h is analytic in B ( a; R ) .
To prove this we use Moreras theorem. Let T be the triangle in B ( a; R ) and ∆ denote inside
of T along with T .
Case 1: When a ∉ ∆ .
Since h is analytic in B ( a; R ) − {a} and T
0 ⇒ T ≈ 0 , by Cauchy theorem ∫ h = 0 .
T
Case 2: When a is vertex of T .
Let T = [a, b, c, a ] be a triangle with a as one of the vertex. For x ∈ [a, b] and y ∈ [a, c] let
P = [ x, b, c, y, x] , then P
0 and by Cauchy theorem ∫ h = 0 .
P
Let T1 = [a, x, y, a ] then ∫ h = ∫ h + ∫ h = ∫ h .
T
T1
P
T1
Since h is continuous and h(a) = 0 , for any ∈> 0 there is δ > 0 such that h( z ) <
∈
for
l (T1 )
z−a <δ .
Now
we
choose
x, y such
that
∈
∫ h( z)dz = ∫ h( z)dz ≤ ∫ h( z ) dz < l (T ) l (T ) .
1
T
T1
T1
1
65
x, y ∈ B(a; δ ) .
Therefore
Hence
∫h = 0.
T
Case 3. When a lies on or inside T
In this case we can construct triangle as shown with belonging to vertex of each constructed
triangle. Therefore using case 2 we must have ∫ h = 0 .
T
Thus
∫ h = 0 for any triangular path T
in B ( a; R ) . Hence by Moreras theorem h is analytic
T
in B ( a; R ) .
y
a
a
a
T
T
x
T
Case 1 : a lies outside T
Case 2 : a is vertex of T
Case 3 : a lies inside T
Corollary 13 An isolated singularity of a function f at z = a is removable singularity iff
f is bounded in the neighborhood of z = a .
Proof. Let z = a is removable singularity of f then there is R > 0 and an analytic function
g : B ( a; R ) →
such that g ( z ) = f ( z ) in 0 < z − a < R .
Since g is continuous at z = a , g (a) is finite. Hence g is bounded in neighborhood of
z = a . Therefore f is bounded in the neighborhood of z = a .
Conversely, suppose f is bounded in the neighborhood of z = a , then there is M > 0 such
that f ( z ) ≤ M in 0 < z − a < δ .
Therefore ( z − a) f ( z ) ≤ z − a M → 0 as z → a .
Thus lim( z − a) f ( z ) = 0 . Hence f has removable singularity at z = a .
z →a
66
Corollary 14 An isolated singularity of a function f
at z = a is removable singularity iff
lim f ( z ) = c .
z →a
Proof. Let z = a is removable singularity at z = a then there is R > 0 and an analytic
function g : B ( a; R ) →
such that g ( z ) = f ( z ) in 0 < z − a < R .
Therefore lim f ( z ) = lim g ( z ) = g (a) = c (say).
z →a
z →a
Conversely, suppose lim f ( z ) = c then lim( z − a) f ( z ) = lim( z − a) lim f ( z ) = 0.c = 0
z →a
z →a
z →a
z →a
Thus lim( z − a) f ( z ) = 0 . Hence f has removable singularity at z = a .
z →a
Definition 15 An isolated singularity at z = a of a function f is pole if lim f ( z ) = ∞ .
z →a
Ex. f ( z ) =
1
has pole at z = 0 .
z
Ex. f ( z ) =
z2
has pole at z = 1, i .
( z − 1)( z − i) 2
Theorem 16 If G is a region with a in G and if f is analytic on G − {a} with pole at z = a ,
then there is a positive integer m and an analytic function g : G →
f ( z) =
such that
g ( z)
.
( z − a) m
Proof. Suppose z = a is pole of f then lim f ( z ) = ∞ . Therefore lim
z →a
Then
z →a
1
= 0.
f ( z)
1
has removable singularity at z = a . Then there is an analytic function
f ( z)
h : B ( a; R ) →
such that h( z ) =
1
when 0 < z − a < R .
f ( z)
 1
if z ≠ a

Now we define h( z ) =  f ( z )
0
if z = a

67
Since h(a) = 0 , there is an analytic function h1 such that h( z ) = ( z − a)m h1 ( z ) and h1 (a) ≠ 0
for some integer m ≥ 1 .
1
= ( z − a ) m h1 ( z )
f ( z)
For z ≠ a ,
Therefore
1
= ( z − a)m f ( z )
h1 ( z )
1
1
=
<∞
z →a h ( z )
h1 (a )
1
Hence lim( z − a ) m f ( z ) = lim
z →a
Then ( z − a ) m f ( z ) has removable singularity at z = a .
By definition there is an analytic function g : B ( a; R ) →
such that g ( z ) = ( z − a ) m f ( z ) ,
z ≠ a.
Hence f ( z ) =
g ( z)
, z ≠ a.
( z − a) m
Definition 17 If f
has pole at z = a and m is smallest positive integer such that
( z − a ) m f ( z ) has removable singularity at z = a , then f has a pole of order m at z = a .
Ex. f ( z ) =
ez
has pole of order 2.
(1 − cos z )
Definition 18 An isolated singularity at z = a of a function f is essential singularity if it is
neither pole nor removable singularity.
1
Ex. f ( z ) = e z has essential singularity at z = 0 .
1
Ex. f ( z ) = sin   has essential singularity at z = 0 .
z
Theorem 19 Casorati-Weierstrass Theorem If f has an essential singularity at z = a ,
then for every δ > 0 f ( ann ( a;0, δ ) ) =
, that is f ( ann ( a;0, δ ) ) is dense in
68
.
Proof. Suppose z = a is essential singularity of f , then we have to prove that for any
c ∈ and δ > 0 , c is the limit point of f ( ann ( a;0, δ ) ) . In another words we have to prove
that, for any given ∈, δ > 0 , there is z ∈ ann ( a;0, δ ) such that
f ( z ) − c <∈ .
On contrary suppose the theorem is false. Then there exists ∈> 0 such that for any δ > 0 ,
f ( z ) − c ≥∈ , for all z ∈ ann ( a;0, δ ) .
Hence lim
z →a
pole, then
f ( z) − c
f ( z) − c
= ∞ . Therefore
has a pole at z = a . Let m be the order of
z −a
z−a
( z − a)
m
( f ( z) − c )
( z − a)
has removable singularity at z = a .

m ( f ( z) − c ) 
Therefore lim ( z − a ) ( z − a )
=0
z →a
( z − a) 

That is
lim ( z − a )
m
z →a
( f ( z) − c ) = 0
Therefore lim ( z − a ) f ( z ) = lim ( z − a )
m
z →a
m
z →a
= lim ( z − a )
z →a
m
( f ( z) − c + c )
( f ( z ) − c ) + lim
( z − a)
z →a
m
c
=0
Thus f will have either removable singularity or a pole , which is a contradiction.
Hence the theorem.
69
EXERCISES
1)
Discuss the singularities of the following functions:
a) f ( z ) =
2)
b) f ( z ) =
cos( z )
z2 +1
c) f ( z ) =
z ( z − 1)
z
d) f ( z ) =
log( z + 1)
.
z2
Discuss the singularities of the following function and classify them
z +i
a) 2
z +1
3)
sin( z )
z
1
b) 2
( z + 1)( z − 6)
1
c) ze
z2
Prove that an entire function has a pole at ∞ of order m if and only if it is a
polynomial of degree m.
4)
Prove that an entire function has removable singularity if and only if it is a constant
function.
70
UNIT - VI
OPEN MAPPING THEOREM AND GOURSAT
THEOREM
In this unit we shall discuss that zeros of an analytic function which are used to
evaluate some complex integrals. We prove open mapping theorem and prove that a
differentiable complex valued function defined on an open set is analytic on the set.
Definition 1 Let γ 1 : [ 0,1] →
, γ 2 : [ 0,1] →
be two closed rectifiable curves in a region
G ; then γ 1 is homotopic to γ 2 in G , written as γ 1
γ 2 , if there is a continuous function
Γ : [ 0,1] × [ 0,1] → G such that
( 0 ≤ s ≤ 1)
1) Γ( s, 0) = γ 1 ( s ) and Γ( s,1) = γ 2 ( s )
2) Γ(0, t ) = Γ(1, t )
( 0 ≤ t ≤ 1) .
Definition 2 A closed rectifiable curve γ is homotopic to zero, if γ is homotopic to a
constant curve and is written as γ
Definition 3 A closed
w∈
0.
rectifiable curve γ is homologous
to zero, if n(γ , w) = 0 for
− G and is written as γ ≈ 0 .
Theorem 4 Let G be a region and let f : G →
be an analytic function on G with zeros
a1 ,..., am (repeated according to multiplicity). If γ is a closed rectifiable curve in G which
does not pass through any point ak and if γ ≈ 0 then
m
1
f '( z )
dz
=
n(γ ; ak ) .
∑
2π i ∫γ f ( z )
k =1
71
Proof. Since
a1 ,..., am are zeros of
f , there exists an analytic function g such that
f ( z ) = ( z − a1 )...( z − am ).g ( z ) , where g (ak ) ≠ 0 for k = 1,..., m and that, g is non-vanishing
on G .
Taking logarithmic differentiation on both sides we get
f '( z )
1
1
g '( z )
=
+ ... +
+
f ( z ) ( z − a1 )
( z − am ) g ( z )
Therefore
m
1
f '( z )
1
1
g '( z )
dz
=
dz + ∫
dz
∑
∫
∫
2π i γ f ( z )
k =1 2π i γ ( z − ak )
γ g ( z)
Since g is non-vanishing
g '( z )
is analytic in G . Hence
g ( z)
g '( z )
∫γ g ( z ) dz = 0 .
Thus
m
1
f '( z )
dz
=
n(γ ; ak ) .
∑
2π i ∫γ f ( z )
k =1
Corollary 5 Let G be a region and let f : G →
be an analytic function on G such that
a1 ,..., am satisfies f ( z ) = α (repeated according to multiplicity). If γ is a closed rectifiable
curve in G which does not pass through any point ak and if γ ≈ 0 then
m
1
f '( z )
dz
=
n(γ ; ak ) .
∑
2π i ∫γ f ( z ) − α
k =1
Proof. Let F ( z ) = f ( z ) − α , which is analytic and a1 ,..., am are zeros of F .
Therefore by theorem
m
1 F '( z )
dz
=
n(γ ; ak )
∑
2π i ∫γ F ( z )
k =1
Thus
m
1
f '( z )
dz
=
n(γ ; ak ) .
∑
2π i ∫γ f ( z ) − α
k =1
72
2z + 1
dz .
z + z +1
z =2
∫
Example 6 Evaluate
2
Solution. Let f ( z ) = z 2 + z + 1 , f '( z ) = 2 z + 1 .
w1 =
Here zeros of f
−1 + i 3
−1 − i 3
and w2 =
are lies inside z = 2
2
2
Then n(γ ; w1 ) = 1 and n(γ ; w2 ) = 1 .
Therefore
z =2
2
f '( z )
dz = ∑ n(γ ; ak )
f ( z)
k =1
1
2π i
2z +1
dz = 1 + 1
z + z +1
z =2
1
2π i
∫
∫
2
2z +1
dz = 4π i .
z
z
1
+
+
z =2
∫
2
Theorem 7 Suppose f is analytic B(a, R ) in and let α = f (a ) . If
f ( z ) − α has a zero of
order m at z = a then there is an ∈> 0 and δ > 0 such that for ζ − a < δ ; the equation
f ( z ) = ζ has exactly m simple roots in B(a,∈) .
Proof. Let F ( z ) = f ( z ) − α , then f is analytic and z = a is zero of order m in B(a, R ) .
Since
zeros
of
analytic
F ( z ) = f ( z ) − α ≠ 0 in
functions
are
isolated,
there
is
0 <∈< R such
2
that
0 < z − a < 2 ∈ and f '( z ) ≠ 0 in 0 < z − a < 2 ∈ .
Let γ (t ) = a + ∈ e2π it , ( 0 ≤ t ≤ 2π ) .
Let σ = foγ
then σ
is a closed rectifiable curve in
f ( B(a, R) ) . Since a ∉ {γ } ,
α = f (a ) ∉ {σ } , then there is δ > 0 such that B(α , δ ) ∩ {σ } = φ . Thus B(α , δ ) lies in the
same
component
of
− {σ } .
Therefore
n(σ ; α ) = n(σ ; ζ ) .
73
for
any
ζ such
that
0 < α −ζ < δ ,
Therefore
1
1
1
1
dw =
dw
∫
∫
2π i σ w − α
2π i σ w − ζ
1
f '( z )
1
f '( z )
dz =
dz
∫
∫
2π i γ f ( z ) − α
2π i γ f ( z ) − ζ
Therefore
∑ n (γ ; z (α ) ) = ∑ n (γ ; z (ζ ) )
k
k
k
where zk (α ) and zk (ζ ) are zeros of f ( z ) − α
k
and f ( z ) − ζ respectively.
Since z = a is zero of order m of f ( z ) − α , zk (α ) = a for each k ,
∑ n (γ ; z (α ) ) = m .
k
k
Therefore
∑ n (γ ; z (ζ ) ) = m .
k
k
Since n ( γ ; zk (ζ ) ) must be either 0 or 1, there are exactly m zeros of f ( z ) = ζ in B (a,∈) .
Since f '( z ) ≠ 0 in 0 < z − a <∈ , all these zeros must be simple in 0 < z − a <∈ .
Thus f ( z ) = ζ has exactly m zeros in B(a,∈) for any ζ ∈ B(α ; δ ) − {σ } .
We now prove open mapping theorem:
Theorem 8 Let G be a region and suppose that f is a non-constant analytic function on G .
Then for any open set U in G , f (U ) is open.
Proof. Let α = f (a ) ∈ f (U ) for some a ∈ U .
Since U is open, there is R > 0 such that B (a, R ) ⊆ U and f ( B(a, R) ) ⊆ f (U ) .
Since f is analytic on G , f is analytic on B(a, R ) . Hence there exists ∈> 0 and δ > 0
such that B(a,∈) ⊆ B(a, R) and that B(α , δ ) ⊆ f ( B(a,∈) ) ⊆ f ( B(a, R) ) ⊆ f (U ) .
Hence for any α ∈ f (U ) , there is δ > 0 such that B(α , δ ) ⊆ f (U ) .
Therefore f (U ) is open.
Theorem 9 Goursat’s Theorem Let G be an open set and
function, then f is analytic.
74
f :G →
be differentiable
Proof. Let f : G →
be differentiable function, then f is continuous function on G .
To prove that f is analytic we shall use Morera’s theorem , it is sufficient to prove that
∫f
= 0 for every triangular path T in G .
T
Let T = [a, b, c, a ] be a triangular path in G . Let ∆ denote the inside of T along with its
boundary.
Let d = diameter (∆ ) and l = length (T ) .
Let T1 , T2 , T3 , T4 be four triangle formed by midpoints of sides of T .
Then we have
∫ f =∫ f +∫ f +∫ f +∫ f
T
T1
T2
T3
…(1)
T4
Let T (1) denote a triangle amongst T1 , T2 , T3 , T4 such that,
∫f
Tj
Then
∫
≤
( j = 1, 2,3, 4)
f
T (1)
∫f
≤4
∫
…(2)
f
T (1)
T
Let ∆ (1) denote inside of T (1) along with its boundary.
Then diam (∆ (1) ) =
1
1
d and l (T (1) ) = l .
2
2
Repeating above process we obtain sequence of triangular paths T (1) , T (2) ,..., T ( n ) such that
∫
T
∫
f ≤4
( n−1)
T
…(3)
f
(n)
and closed sets ∆ (1) , ∆ (2) ,...∆ ( n ) such that ∆ (1) ⊇ ∆ (2) ⊇ ...
diam (∆ ( n ) ) =
1
1
d and l (T ( n ) ) = n l .
n
2
2
75
Therefore
∫f
∫
≤ 4n
T
T
f
(n)
Since ∆ (1) ⊇ ∆ (2) ⊇ ... is a descending series of closed sets such that, diam (∆ ( n ) ) =
as n → ∞ .
Therefore by Cantor’s theorem, ∩ ∆ ( n ) = {z0 } for some z0 ∈ G .
n
Since f is differentiable in G , f is differentiable at z0 .
Hence , given ∈> 0 there is δ > 0 such that,
f ( z ) − f ( z0 )
− f '( z0 ) <∈ whenever 0 < z − z0 < δ
z − z0
Equivalently, f ( z ) − f ( z0 ) − ( z − z0 ) f '( z0 ) <∈ z − z0 , 0 < z − z0 < δ .
Since diam (∆ ( n ) ) =
1
d → 0 as n → ∞ , there is n0 > 0 such that diam (∆ ( n ) ) < δ .
2n
Since z0 ∈ ∆ ( n ) for each n , z0 ∈ ∆ ( n0 ) also. Hence ∆ ( n0 ) ⊆ B ( z0 ; δ ) .
Hence for n ≥ n0
∫
T
T
∫
≤
 f ( z ) − f ( z0 ) − ( z − z0 ) f '( z0 ) dz
∫
f =
(n)
(n)
f ( z ) − f ( z0 ) − ( z − z0 ) f '( z0 ) dz
T (n)
∫ ∈ z−z
≤
T
0
dz
(n)
≤∈ diam (∆ ( n ) ) l (T ( n ) ) =∈
Thus
∫f
T
Therefore
∫
≤ 4n
T
∫f
dl
4n
f ≤∈ d l
n ≥ n0 ,
(n)
= 0.
T
76
1
d →0
2n
EXERCISE
1)
Let p(z) be a polynomial of degree n and let R > 0 be sufficiently large so that p never
vanishes in {z : | z |≥ R} . If γ (t ) = Reit , 0 ≤ t ≤ 2π , show that
∫γ
p' ( z)
dz = 2π in .
p( z )
77
UNIT - VII
LAURENT SERIES DEVELOPMENT AND RESIDUE
THEOREM
In this unit we shall discuss Laurent’s series expansion of complex valued functions
and prove residue theorem.
∞
Definition 1 A series of the form
∑ a ( z − a)
n
n
is called double series about z = a .
n =−∞
∞
Definition 2 The double series
∑ a ( z − a)
n
n
is said to be absolutely convergent, if both the
n =−∞
∞
series
∑ an ( z − a) and
n
n =0
−∞
∑ a ( z − a)
n
n
are absolutely convergent.
n =−1
∞
Definition 3 The double series
∑ a ( z − a)
n
n
is said to be uniformly convergent, if both the
n =−∞
∞
series
∑ an ( z − a)n and
n =0
−∞
∑ a ( z − a)
n
n
are uniformly convergent.
n =−1
Theorem 4 Let f be analytic in annulus ann ( a; R1 , R2 ) . Then
f ( z) =
∞
∑ a ( z − a)
n
…(1)
n
n =−∞
where the convergence is absolute and uniform over ann ( a; r1 , r2 ) if R1 < r1 < r2 < R2 .
Also coefficients an are given by the formula
an =
1
f ( z)
dz
∫
2π i γ ( z − a ) n +1
…(2) where γ is
the circle z − a = r for any r , R1 < r < R2 . Moreover the series is unique.
78
Proof. We shall begin by showing that the integral in (2) is independent of r , so that for each
integer n , an is constant.
Let R1 < r1 < r2 < R2 and γ 1 = a + r1eit , γ 2 = a + r2 eit ( 0 ≤ t ≤ 2π ).
Let γ = γ 2 + [ z2 , z1 ] − r1 + [ z1 , z2 ] then γ
by Cauchy’s theorem we have
Therefore
f ( w)
∫
γ (w − a)
n +1
dw +
2
Thus
f ( w)
∫
γ (w − a)
n +1
dw =
2
0 . Since f be analytic in annulus ann ( a; R1 , R2 ) ,
f ( w)
∫γ (w − a)
n +1
dw = 0 .
f ( w)
f ( w)
f ( w)
dw + ∫
dw + ∫
dw = 0
n +1
n +1
n +1
w
a
w
a
w
a
(
−
)
(
−
)
(
−
)
[ z2 , z1 ]
− γ1
[ z1 , z2 ]
∫
f ( w)
∫
γ (w − a)
n +1
dw .
1
Now let z ∈ ann ( a; r1 , r2 ) then by Cauchy integral formula we have,
f ( z) =
1
f ( w)
1
f ( w)
dw −
dw
∫
∫
2π i γ 2 ( w − z )
2π i γ 1 ( w − z )
We define
f1 ( z ) =
−1
f ( w)
1
f ( w)
dw and f 2 ( z ) =
dw .
∫
∫
2π i γ1 ( w − z )
2π i γ 2 ( w − z )
Therefore, f ( z ) = f1 ( z ) + f 2 ( z ) .
…(3)
Now
f2 ( z) =
=
1
f ( w)
1
f ( w)
dw =
dw
∫
∫
2π i γ 2 ( w − z )
2π i γ 2 ( w − a ) − ( z − a )
1
2π i γ∫2
f ( w)
dw
 ( z − a) 
( w − a ) 1 −

 ( w − a) 
n
1
f ( w) ∞  z − a 
=
∑
 dw
2π i γ∫2 ( w − a ) n = 0  w − a 
z−a
< 1 on γ 2
w−a
79
∞
 1

f ( w)
=∑ 
dw ( z − a) n
n +1
∫
n =0 
 2π i γ 2 ( w − a )

∞
= ∑ an ( z − a ) n .
…(4)
n =0
1
f ( w)
dw .
∫
2π i γ 2 ( w − z )n +1
where an =
Also
f1 ( z ) =
=
−1
f ( w)
1
f ( w)
dw =
dw
∫
∫
2π i γ1 ( w − z )
2π i γ1 ( z − a ) − ( w − a )
1
2π i γ∫1
f ( w)
dw
 (w − a ) 
( z − a ) 1 −

 ( z − a) 
m
=
1
f ( w) ∞  w − a 
∑
 dw
2π i γ∫1 ( z − a ) m = 0  z − a 
w−a
< 1 on γ 1
z −a
∞
 1

m
− m −1
=∑ 
−
(
w
a
)
f
(
w
)
dw
 ( z − a)
∫
m =0 
 2π i γ 1

∞
= ∑ bm ( z − a ) − m −1 .
m =0
where bm =
=
1
( w − a ) m f ( w)dw .
2π i γ∫1
1
f ( w)
dw
∫
2π i γ1 ( w − a ) − m
80
=
1
f ( w)
dw
∫
2π i γ1 ( w − a )( − m −1) +1
= a− m −1
∞
Therefore f1 ( z ) = ∑ a− m −1 ( z − a ) − m −1
m=0
=
−∞
∑ a ( z − a)
n
...(5)
n
n =−1
Therefore from (3), (4) and (5) we have
f ( z) =
−∞
∞
∑ an ( z − a ) n + ∑ an ( z − a ) n =
n =−1
n=0
∞
∑ a ( z − a)
n
n
for z ∈ ann ( a; r1 , r2 ) .
n =−∞
Uniqueness:
f ( z) =
Let
∞
∑ c ( z − a)
n
n
be another Laurent series of f where
n =−∞
an =
1
f ( z)
dz
∫
2π i γ ( z − a ) n +1
∞
ck ( z − a ) k
1 k∑
=−∞
=
dz
2π i γ∫1 (z − a)n +1
=
1 ∞
ck ∫ ( z − a )k − n −1 dz
∑
2π i k =−∞ γ1

2π i if n = −1
 ∫ ( z − a)n dz = 

 z − a = r
 0 if n ≠ − 1
= cn
Hence the uniqueness.
z=a
Corollary 5 Let
f ( z) =
∞
∑ a ( z − a)
n
n
be an isolated singularity of a function
be its Laurent series expansion in ann ( a; 0, R ) . Then :
n =−∞
81
f
and let
a) z = a is a removable singularity iff an = 0 for n ≤ −1 .
b) z = a is a pole of order m iff a− m ≠ 0 and an = 0 for n ≤ −(m + 1) .
c) z = a is an essential singularity iff an ≠ 0 for infinitely many negative integers.
Proof. a) Suppose z = a is removable singularity of f then there is R > 0 and an analytic
function g : B ( a; R ) →
such that g ( z ) = f ( z ) in 0 < z − a < R . Since g is analytic in
∞
B ( a; R ) by Taylor’s theorem g ( z ) = ∑ bn ( z − a ) n for z − a < R .
n =0
∞
Therefore f ( z ) = g ( z ) = ∑ bn ( z − a)n for 0 < z − a < R .
n =0
This is Laurent series expansion of f in ann ( a; 0, R ) then by uniqueness of Laurent series
we must have an = bn for all n .
Therefore, an = 0 for n ≤ −1 .
Conversely, suppose that an = 0 for n ≤ −1 then Laurent series expansion of f
is
∞
f ( z ) = ∑ an ( z − a ) n = a0 + a1 ( z − a ) + a2 ( z − a )2 + ... + .
n =0
∞

Therefore, lim( z − a) f ( z ) = lim( z − a )  ∑ an ( z − a) n  = 0.a0 = 0 .
z →a
z →a
 n=0

Thus z = a is removable singularity of f .
b) Suppose z = a is a pole of order m then ( z − a ) m f ( z ) has removable singularity at z = a .
∞
Therefore ( z − a ) m f ( z ) = ( z − a )m ∑ an ( z − a )n
n =0
∞
= ∑ an ( z − a ) m + n
n =0
From part (a) an = 0 for all m + n ≤ −1 .
That is a− m ≠ 0 and an = 0 for n ≤ −(m + 1) .
Conversely, suppose a− m ≠ 0 and an = 0 for n ≤ −(m + 1) then Laurent series expansion of f
is
f ( z) =
∞
∑ a ( z − a)
n
n
= a− m ( z − a ) − m + ... + a−1 ( z − a ) −1 + a0 + a1 ( z − a )1 + ... .
n =− m
82
Then ( z − a )m f ( z ) = a− m + ... + a−1 ( z − a ) m −1 + a0 ( z − a )m + a1 ( z − a ) m +1 + ... .
Thus ( z − a ) m f ( z ) has Laurent series expansion which does not contains negative powers of
( z − a) .Therefore from part (a) ( z − a ) m f ( z ) has removable singularity at z = a .
c) (a) and (b) together implies (c).
1
Example 6 Find Laurent series expansion of f ( z ) = e z at z = 0 .
Solution: we have e z = 1 + z +
z 2 z3
+ + ...
2! 3!
Therefore,
2
3
1 1
1
1 z z
e z = 1 + +   +   + ... if z > 0 .
z
2!
3!
1
1
1
= 1+ +
+
+ ... .
2
z 2! z 3! z 3
1
∞
1
for ann ( 0;0, ∞ ) .
n
n=0 n ! z
ez = ∑
1
in
z ( z − 1)( z − 2)
Ex ample 7 Find Laurent series expansion of
a) ann ( 0;0,1)
b) ann ( 0; 2, ∞ )
Solution: Let f ( z ) =
1
.
z ( z − 1)( z − 2)
a) Consider, ann ( 0;0,1) = { z : 0 < z < 1}
Here z < 1 and
Therefore f ( z ) =
z 1
< < 1.
2 2
1 1
1 
−

z  z − 2 z − 1 
83
=


1
1
1 
+
z  −2 1 − z
(1 − z ) 
2


=
1 1
1− z
2
z  −2
(
)
(
)
1 1 ∞
=  ∑ z
z  −2 n =0 2
−1
−1 
+ (1 − z ) 

n
∞
n

( ) + ∑ ( z )  ,

z < 1 and
n=0
∞
= ∑ 1 − 1 n +1  z n −1 .

2 
n =0
b) ann ( 0; 2, ∞ ) = { z : 2 < z < ∞}
Here z > 2 implies that
Therefore f ( z ) =
1 1
2
< < 1 and
< 1.
z 2
z
1 1
1 
−

z  z − 2 z − 1 
=

1
1
1
−
z  z 1− 2
z 1− 1
z
z

=
1 
1− 2
z
z 2 
=
1 ∞ 2
∑
z 2  n =0 z
(
) (
(
) − (1 − 1 z )
−1
∞
)
−1






( ) − ∑ ( 1 z ) 
n
n=0
∞
= ∑ ( 2n − 1)
n =0
n
1
z
n+ 2
.
84
z 1
< <1
2 2
Definition 8 Let z = a be an isolated singularity of a function f . Then residue of f ( z ) at
1
( that is a−1 ) in Laurent series expansion of
( z − a)
z = a is defined to be coefficient of
f ( z ) about z = a and is denoted by Res ( f ; a ) .
Res ( f ; a ) = a−1 =
1
f ( z ) dz .
2π i ∫γ
 z2 z4 z6

1 − 1 − + − + ... 
2
1 − cos z
 2! 4! 6!
 = 1 − 1 + z − ...
=
Example 9 Let f ( z ) =
z4
z4
2! z 2 4! 6!
Therefore ,
Res ( f ;0 ) = a−1 = coefficient of
1
=0 .
z
We shall now prove residue theorem:
Theorem 10 Let f be analytic in a region G except for isolated singularities a1, a2 ,.., am . If
γ is a close rectifiable curve in G which does not pass through any of the points ak and if
γ ≈ 0 in G then
m
1
f
(
z
)
dz
=
n ( γ ; ak ).Res ( f ; ak )
∑
2π i ∫γ
k =1
mk = n ( γ ; ak ) , (1 ≤ k ≤ m ) . Let r1, r2 ,.., rm be positive numbers such that no two
Proof. Let
disks B ( ak ; rk ) intersect, none of them intersects γ and each disk is contained in G . Let
γ k ( t ) = ak + rk e−2π im t
k
0 ≤ t ≤1.
Then for each k = 1, 2,..., m ,
n ( γ ; ak ) + n ( γ k ; ak ) = mk − mk = 0 and that n ( γ j ; ak ) = 0 for k ≠ j .
Therefore
m
n ( γ ; ak ) + ∑ n ( γ j ; ak ) = 0
k = 1, 2,..., m .
…(1)
j =1
Since γ ≈ 0 , n ( γ ; w ) = 0,
w ∉ G also n ( γ k ; w ) = 0,
w∉G .
Therefore for w ∉ G
m
n ( γ ; w) + ∑ n (γ j ; w ) = 0 .
…(2)
j =1
85
Thus from (1) and (2) we have
m
n ( γ ; w ) + ∑ n ( γ k ; w ) = 0 for w ∉ G − {a1 , a2 ,..., am } .
k =1
Since f be analytic in G − {a1 , a2 ,..., am } , by Cauchy’s theorem we have
∫γ
m
f +∑ ∫ f = 0.
k −1 γ k
Now consider
∫
γ
f ( z )dz =
j
j
k
γ k j =−∞
k
where f ( z ) =
∞
∫ ∑ b ( z − a ) dz
∞
∑ b (z − a )
j
j
k
be Laurent series expansion about z = ak
j =−∞
=
∞
∑ b ∫ (z − a )
γ
j
j =−∞
j
k
dz
k
= b−1 ∫ ( z − ak )−1 dz
γk
= Res ( f ; ak ) 2π in ( γ k ; ak )
= − 2π in ( γ ; ak ) .Res ( f ; ak )
Thus
∫γ
m
f ( z ) + ∑ − 2π i n ( γ ; ak ) .Res ( f ; ak ) = 0 .
k −1
Hence,
m
1
f
(
z
)
dz
=
n ( γ ; ak ).Res ( f ; ak ) .
∑
2π i ∫γ
k =1
86
G
γ
a1
a2
a3
Theorem 11 Suppose f has a pole of order m at z = a and g ( z ) = ( z − a) m f ( z ) then
Res ( f ; a ) =
1
m −1
g( ) (a)
( m − 1)!
Proof. Since f has a pole of order m at z = a , g ( z ) = ( z − a) m f ( z ) has removable
∞
singularity at z = a . Then g has Laurent series expansion of the form g ( z ) = ∑ bn ( z − a ) n
n =0
where bn =
1 (n)
g ( a ) , in 0 < z − a < R for some R > 0 .
( n )!
Therefore, f ( z ) =
b0
( z − a)
Thus Res ( f ; a ) = bm −1 =
m
+
b1
( z − a)
m −1
∞
bm −1
+ ... +
+ ∑ bn ( z − a) n − m .
( z − a ) n=m
1
m −1
g( ) (a) .
( m − 1)!
Corollary 12 If f has a simple pole at z = a then Res ( f ; a ) = g ( a ) = lim ( z − a ) f ( z ) .
z →a
Proof. We have Res ( f ; a ) =
1
1−1
g( ) (a)
(1 − 1) !
= g (a)
= lim g ( z )
z →a
= lim ( z − a ) f ( z ) .
z →a
87
Corollary 13 If f has a simple pole at z = a and f ( z ) =
h( z )
h(a )
then Res ( f ; a ) =
.
k ( z)
k '(a )
Proof. Since f has a simple pole at z = a , k has simple zero at z = a , k (a) = 0 .
So that, k ( z ) = ( z − a) g ( z ) , where g is analytic and non-vanishing at z = a . Moreover,
k '(a) = g (a ) Thus
Res ( f ; a ) = lim ( z − a ) f ( z )
z →a
= lim ( z − a )
z →a
h( z )
k ( z)
= lim h( z ) lim
z →a
z →a
= h(a ).
( z − a)
( z − a) g ( z )
1
h( a )
=
.
g (a ) k '(a )
z2
Example 14 Calculate residue of
.
( z − 1)( z − 2) 2
Sol. Let f ( z ) =
z2
then f has simple pole at z = 1 and pole of order 2 at z = 2 .
( z − 1)( z − 2)2
a) Res ( f ;1) = lim ( z − 1) f ( z )
z →1
= lim ( z − 1)
z →1
z2
( z − 1)( z − 2)2
=1
Alternatively, Res ( f ;1) =
h(1)
where h( z ) = z 2 , k ( z ) = ( z − 1)( z − 2) 2 = z 3 − 5 z 2 + 8 z − 4
k '(1)
and k '( z ) = 3 z 2 − 10 z + 8
Therefore Res ( f ;1) =
12
h(1)
= 2
= 1.
k '(1) 3.1 − 10.1 + 8
88
z2
b) z = 2 is pole of order 2. Let g ( z ) = ( z − 2) f ( z ) =
then
( z − 1)
2
g '( z ) =
( z − 1).2 z − 1.z 2 z 2 − 2.z
=
( z − 1) 2
( z − 1)2
Therefore
Res ( f ; 2 ) =
1
2 −1
g ( ) ( 2)
( 2 − 1)!
= g ' ( 2)
= 0.
Ex ample 15 Calculate residue of
Sol. Let f ( z ) =
z +1
.
z − 2z
2
z +1
h( z )
=
then f has two simple polesnamely, at z = 0 and z = 2 .
z − 2 z k ( z)
2
Res ( f ;0 ) =
h(0)
0 +1
1
=
= − and
2
k '(0) 2.0 − 2
Res ( f ; 2 ) =
h(2)
2 +1
3
=
= .
k '(2) 2.2 − 2 2
Evaluation of real integrals using residue theorem:
We shall now discuss the methods of evaluating real integrals using residue calculus.
∞
Type-I : Integration of the type
∫
f ( x)dx where f ( x) =
−∞
polynomials in x and deg ( g ( x) ) − deg ( h( x) ) ≥ 2 then
∫ f ( z )dz → 0
C
is the semicircle z = R in the upper half of the plane.
∞
Example 16 Evaluate
1
∫ 1+ x
2
dx .
0
∞
∞
1
1
1
Solution. Let I = ∫
dx = ∫
dx .
2
1+ x
2 −∞ 1 + x 2
0
89
h( x )
g ( x)
and h( x) , g ( x) are
as R → ∞ , where C
Let f ( z ) =
1
and γ = [− R, R] ∪ C where C is the semicircle z = R lie in the upper half
1+ z2
of the plane. Here we choose R so that all pole in the upper half of the plane are in the
interior of γ .
Here poles of f ( z ) =
1
are z = i, −i and z = i lies in upper half of the plane .
1+ z2
Since z = i = 1 , the pole z = i lies inside γ if we choose R > 1 .
Therefore by residue theorem
1
∫γ f ( z) dz = ∫γ 1 + z
2
dz = 2π i.Res ( f ;i ) .
Since z = i is a simple pole of f ( z )
Res ( f ;i ) = lim( z − i )
z →i
1
1
=
2
1+ z
2i
1
Thus
∫γ f ( z) dz = ∫γ 1 + z
Also
∫γ
R
f ( z ) dz =
R
π=
∫
−R
1
∫ 1+ x
−R
2
2
dz = 2π i.
1
=π
2i
…(1)
f ( x)dx + ∫ f ( z )dz
C
1
dz
2
1
z
+
C
dx + ∫
…(2)
Consider,
∫
C
π
π
1
R eit
it
f ( z )dz = ∫
R
idt
=
i
e
∫0 1 + R 2e2it dt .
1 + R 2 e2 it
0
Since,
|1 + z 2 |≥||1| − | z 2 ||=|1 − R 2 |= R 2 − 1 ,
we have
∫
C
π
f ( z )dz ≤ ∫
0
R
dt .
R −1
2
90
Since
1
R
→ 0 , as R → ∞ , we get lim ∫
dz = 0
R →∞ 1 + z 2
R −1
C
2
On taking limit as R → ∞ in (2), we obtain
∞
π=
1
∫ 1+ x
2
dx + 0 .
−∞
∞
Therefore,
∞
1
1
1
π
∫0 1 + x 2 dx = 2 −∞∫ 1 + x 2 dx = 2 .
∞
Example 17 Evaluate
∞
Solution. Let I =
x2
∫ 1 + x 4 dx .
−∞
1
∫ 1+ x
2
dx .
−∞
z2
Let f ( z ) =
and γ = [− R, R ] ∪ C where C is the semicircle z = R lie in the upper half
1+ z4
of the plane. Here we choose R so that all pole lies in the upper half of the plane are in the
interior of γ .
πi
3π i
5π i
7π i
πi
3π i
z2
4
4
4
4
4
Here poles of f ( z ) =
are z = e , e , e , e
and z = e , e 4 are lies in upper half
1+ z4
of the plane and inside γ if we choose R > 1 .
Therefore by residue theorem
 π4i 
h
e 
πi



 =
Res  f ;e 4  =
πi

 g ' e 4 
 
 
∫γ
πi
3π i





z2
4
4
=
2
π
.
Res
;
+
Res
;
f ( z ) dz = ∫
dz
i
f
e
f
e

.




4



 
γ 1+ z

2
 π4i 
e 
πi
  = 1 e − 4 = 1 cos  π i  − i sin  π i   = (1 − i )
 
 
3
4
4   4 
 4  4 2
 π4i 
4e 
 
and

Res 

 3π4 i 
h
e 
3π i


 =
f ;e 4  =
3π i
 g ' e 4 




2
 3π4 i 
e 
3π i

 = 1 e − 4 = 1 cos  3π i  − i sin  3π i   = (−1 − i )




3
4
4   4 
4 2
 4 
 3π4 i 
4e 


91
Thus
∫γ
z2
 (1 − i ) (−1 − i )  π
f ( z ) dz = ∫
dz = 2π i. 
+
=
4
4 2 
2
4 2
γ 1+ z
Also
∫γ
f ( z ) dz =
π
R
∫
−R
f ( x)dx + ∫ f ( z )dz
C
R
2
=
x2
z2
dx
+
∫− R 1 + x 4 C∫ 1 + z 4 dz
…(2)
Consider,
∫
C
π
π
R 2 e 2it
R 3 e3it
it
f ( z )dz = ∫
R e idt = i ∫
dt .
1 + R 4 e4 it
1 + R 4 e 4it
0
0
Since,
|1 + z 4 |≥||1| − | z 4 ||=|1 − R 4 |= R 4 − 1 ,
we have
∫
C
…(1)
π
f ( z )dz ≤ ∫
Since
0
R3
dt .
R4 −1
R3
z2
dz = 0
,
as
R
→
0
→
∞
,
we
get
lim
R →∞ ∫ 1 + z 4
R4 −1
C
On taking limit as R → ∞ in (2) we get
π
2
∞
=
x2
∫−∞ 1 + x 4 dx + 0 .
∞
Therefore,
x2
π
∫−∞ 1 + x 4 dx = 2 .
92
2π
Type-II : Integration of the type
∫
f (cos θ ,sin θ )dθ where f (cos θ ,sin θ ) is rational
0
function of cos θ and sin θ .
Here we substitute z = eiθ
( 0 ≤ θ ≤ 2π )
that is z = eiθ describes the unit circle z = 1 .
eiθ + e − iθ z + z −1 z 2 + 1
eiθ − e − iθ z − z −1 z 2 − 1
Also cos θ =
=
=
and sin θ =
=
=
.
2
2
2z
2i
2i
2iz
2π
Then
∫
0
 z 2 + 1 z 2 − 1  dz
,
f (cos θ ,sin θ )dθ = ∫ f 
where γ is positively oriented unit circle

2iz  iz
 2z
γ
z = 1 can be evaluated using residue theorem.
2π
Example 18 Evaluate
1
∫ 1 + a sin θ dθ ( −1 < a < 1) .
0
2π
Solution. Let I =
1
∫ 1 + a sin θ dθ .
…(1)
0
( 0 ≤ θ ≤ 2π )
Put z = eiθ
2π
Therefore
then dz = izdθ and sin θ =
1
∫ 1 + a sin θ dθ = ∫γ
0
=
Let f ( z ) =
z2 −1
.
2iz
1
dz
where γ is unit circle z = 1 .
2
 z − 1  iz
1+ a 

 2iz 
2
1
dz
∫
a γ  2 2i

 z + z − 1
a


…(2)
1
1
−i + i 1 − a 2
−i − i 1 − a 2
and z2 =
=
where z1 =
a
a
 2 2i
 ( z − z1 )( z − z2 )
 z + z − 1
a


are simple poles of f ( z ) .
Note that z2 =
−i − i 1 − a 2 1 + 1 − a 2
=
>1
a
a
( −1 < a < 1) .
93
Since z1 z2 = 1 , z1 < 1 .
Thus the pole z1 lies inside γ and z2 lies outside.
Therefore Res ( f ;z1 ) = lim( z − z1 ) f ( z ) = lim( z − z1 )
z → z1
z → z1
1
1
a
=
=
( z − z1 )( z − z2 ) ( z1 − z2 ) 2i 1 − a 2
Then by residue theorem
1
1
a
dz = Res ( f ;z1 ) .n ( γ ; z1 ) =
∫
2π i γ  2 2i

2i 1 − a 2
 z + z − 1
a


Therefore
1
aπ
dz =
2i

2
1− a2
 z + z − 1
a


∫γ 
Thus from (2) we have
2π
1
2
1
2 aπ
2π
dz =
=
.
2
2
2i
a 1− a

2
1
−
a
 z + z − 1
a


∫ 1 + a sin θ dθ = a ∫γ 
0
π
Example 19 Evaluate
1
∫ a + cos θ dθ ( a > 1) .
0
π
1
1
Solution. Let I = ∫
dθ =
2
a + cos θ
0
Put z = e
iθ
2π
1
∫ a + cos θ dθ .
…(1)
0
z2 +1
.
( 0 ≤ θ ≤ 2π ) then dz = izdθ and cos θ =
2z
1
Therefore I =
2
2π
1
1
∫ a + cos θ dθ = 2 ∫γ
0
1
dz
where γ is unit circle z = 1 .
2
 z + 1  iz
a+

 2z 
94
=
Let f ( z ) =
1
1
dz
∫
2
i γ ( z + 2az + 1)
1
1
=
( z + 2az + 1) ( z − z1 )( z − z2 )
2
…(2)
where z1 = − a + a 2 − 1 and z2 = − a − a 2 − 1
are simple poles of f ( z ) .
Note that z2 = − a − a 2 − 1 = a + a 2 − 1 > 1
( as a > 1) .
Since z1 z2 = 1 , z1 < 1 .
Thus the pole z1 lies inside γ and z2 lies outside.
Therefore Res ( f ;z1 ) = lim( z − z1 ) f ( z ) = lim( z − z1 )
z → z1
z → z1
1
1
1
=
=
( z − z1 )( z − z2 ) ( z1 − z2 ) 2 a 2 − 1
Then by residue theorem
1
1
1
dz = Res ( f ;z1 ) .n ( γ ; z1 ) =
∫
2
2π i γ ( z + 2az + 1)
2 a2 −1
Therefore
∫γ ( z
2
1
πi
dz =
+ 2az + 1)
a2 −1
Thus from (2) we have
π
1
1
∫ a + cos θ dθ == i ∫γ ( z
0
2
1
1 πi
π
dz =
=
.
2
i a −1
+ 2az + 1)
a2 −1
95
EXERCISES
1. Show that
∞
dx
π
i) ∫
=
,
4
1+ x
2
−∞
∞
dx
π
=
iii) ∫
2
4
1+ x + x
2 3
0
∞
xdx
∫ 1+ x
ii)
0
4
=
π
4
,
∞
x 2 dx π
iv) ∫
=
1 + x6 6
0
2. Show that
π /2
i)
∫
0
dx
π
=
,
2
a + sin x 2a a + 1
2π
ii)
iii)
π
0
π
cos xdx
π
∫−π 3 + 4 cos x = − 3
cos 3 xdx
∫ 5 − 4 cos x = 12
π
iv)
cos 2 xdx
∫ 1 − 2a cos x + a
0
2
=
π
12
(a 2 < 1) .
96
UNIT - VIII
ROUCHE’S THEOREM AND MAXIMUM MODULUS
THEOREM
In this unit we shall prove Argument Principle, Rouche’s theorem and Maximum modulus
theorem. Rouche’s theorem is found to be very useful in finding number of zeros inside a
given closed curve.
Definition(Meromorphic function)1. A function f that is analytic except for finite number
of points is meromorphic function.
e.g. i)
1
1
ii)
.
( z − 2)( z + 5)
exp( z ) − 1
Theorem(Argument Principle)2. Let f be meromorhic in G with pole p1, p2, . . . , pm and
zeros z1, z2, . . ., zn counted according to the multiplicity. If γ is a closed rectifiable value in G
with γ ≅ 0 and not passing through p1, p2, . . . , pm and z1, z2, . . ., zn then
1
2π i
∫γ
n
m
f ′( z )
dz = ∑ n(γ j zk ) − ∑ (γ j pk )
f ( z)
k =1
k =1
Proof : Since p1, p2, . . . , pm are poles and z1, z2, . . ., zn are zeros of f, there is an analytic
function g such that


f (z) = 


n
∏
k =1
n
∏
k =1

(z − zk ) 
 ⋅ g (z)
(z − pk ) 

where g is non vanishing.
97
(1)
Logarithmic differention of f gives us
n
f ′( z ) n
1
1
g ′( z )
=∑
−∑
+
f ( z ) k =1 ( z − zk ) k =1 ( z − pk ) g ( z )
(2)
Since g is non vanishing analytic function
g′
∫γ g
= 0 . Hence, (2) implies
n
n
1
f ′( z )
1
dz
1
dz
=
−
+0
dz
∑
∑
∫
∫
∫
2π i γ f ( z )
k =1 2π i γ ( z − z k )
k =1 2π i γ ( z − pk )
n
n
1
f ′( z )
dz
=
n
(
γ
;
z
)
−
n(γ ; pk ) .
∑
∑
k
2π i ∫γ f ( z )
k =1
k =1
Note 3 If γ(t) = a + reit, then
1
f ′( z )
dz = Z f − Pf
∫
2π i f ( z )
where
Zf : Number of zeros of f inside B(a; r) counted according to multiplicity
and
Pf : Number of poles of f inside B(a; r) counted according to multiplicity.
e. g.
Then
f ( z) =
∫γ
( z − 1)3 ( z − 4)4
,
( z + i ) 2 ( z − i ) ( z + 1 − i )3
γ:|z|=2
f ′( z )
dz = 2π i[3 − (2 + 1 + 3)] = – 6πi.
f ( z)
Rouche’s theorem 4 Suppose f and g are meromorphic in a neighborhood of B (a; R) with no
zeros or poles on the circle γ = { z / | z − a |= R} . If Zf ,Zg (Pf, Pg) are the numbers of zeros
(poles) of f & g inside γ counted according to their multiplicities and if | f(z) + g(z) | < | f(z)| +
| g(z)| on {γ}, then Zf – Zg = Pf – Pg or Zf – Pf = Zg – Pg.
98
Proof : If f/g = λ is real, then | f(z) + g(z) | < | f(z)| + | g(z)| gives us
f ( z)
f ( z)
+1 <
+ 1 i.e. | λ + 1 | < | λ | + 1.
g ( z)
g ( z)
Further, if f/g = λ ≥ 0 then (1) gives us, λ + 1 < λ + 1which is absurd.
Hence f / g does not take any value in [0, ∞). Therefore, f / g has well defined logarithm log
(f / g) in C – [0, ∞). Moreover, log (f / g) is primitive of
(f / g)′ / (f / g). Hence
∫γ
f ′( z )
dz = 0 so that
f ( z)
 gf ′ − g ′f   g 
  = 0 ⇒
g2  f 
∫γ 
⇒
1
f′
∫
2π i γ f
=
 f ′ g′ 
− =0
f
g
∫γ 
1 g′
2π i ∫γ g
⇒ Zf – Pf = Zg – Pg.
Corollary 5 If f and g are analytic in neighborhood of B (a ; R) with no zeros on {γ} where γ
: | z – a | = R and if | f(z) + g(z) | < | f(z)|, then f and g have same number of zeros in B(a; R).
Proof : We have | g(z)| < | f(z)| on {γ} ⇒ | f(z) + g(z) | < | f(z)| + | f(z) + g(z) | on {γ}, hence
f and f + g have same number of zeros in B(a; r).
Corollary 6 If f and g are analytic in neighborhood of B (a ; R) with no zeros on {γ} where γ
: | z – a | = R and if | g(z) | < | f(z)| on {γ}, then f and f+g have same number of zeros in B(a;
R).
99
Proof:
We
have
|
g(z)
|
<
|
f(z)|
on
{γ}.
Thus
we
have
| f ( z ) + g ( z ) − f ( z ) |<| f ( z ) | + | f ( z ) + g ( z ) | , hence, f and f + g have same number of
zeros.
Fundamental theorem of Algebra 7 Every non constant poly. has a zero. OR A polynomial
P(z) = z n + a1zn – 1 + . . . + an. has precisely n – zeros.
Proof : Consider
P(z) = z n + a1zn – 1 + . . . + an
⇒ P(z) / z n = 1 + a1/z + . . . + an /zn .
hence P(z) / z n → 1 as n → ∞. Clearly P(z) / z n is analytic in ____ – {0}. Hence is analytic in
the neighborhood of infinity. i.e. {z / | z | ≥ R}.
Let γ : {z / | z | = R}, then
P( z )
− 1 < 1 or | P(z) – zn | < | zn | on {γ} for suitable choice of R.
n
z
Hence Rouche’s theorem, P(z) & zn have same number of zeros in B(a; r). Since zn has n
zeros (counted according to multiplicities) in B(a; r). P(z) also has precisely n- zeros in B(a;
r) for some R > 0.
Example 8 Prove that there are 3 zeros of z3 – 6z + 8 in an B(0; 3).
Proof : Let f1(z) = z3 – 6z + 8, We shall prove that f1 has 3 zeros in B(0; 3)and no zeros in
B(0; 1).
Let f(z) = z3 and g(z) = – 6z + 8
Now, on | z | = 3, | f(z) | = | z |3 = 27, and | g(z) | = | – 6z + 8 | ≤ 6 | z| + 8 = 26.
Thus | g(z) | ≤ 26 < 27 = | f(z)| on | z | = 3.
Hence f and f + g have same nos. of zeros in B(0; 3). Since f(z) = z3 has 3 zeros in B(0; 3). f
+ g = z3 – 6z + 8 also has 3 zeros in B(0; 3).
100
Now consider | z | = 1 .
Let f(z) = z3 + 8 and g(z) = – 6z .
Now | f(z)| = | z3 + 8 | ≥ 8 – | z |3 = 8 – 1 = 7.
& | g(z) | = | – 6z | = 6 | z| = 6.
Thus | f(z)| = 6 < 7 ≤ | f(z)| on | z | = 1.
Hence f and f + g have same number of zeros in B(0; 3).
Since f(z) = z3 + 8 has 3 zeros on | z | = 2, f has no zeros in B(0; 1).
Hence f(z)+ g(z) = z3 – 6z + 8 has no zeros in B(0; 1).
Thus all the three zeros of z3 – 6z + 8 lie in of ann(0; 1,3).
Example 9 Let > 1, and show that the equation of λ – z – e–z = 0 has exactly one solution in
the half plane { z / Re(z) > 0 }. Prove that this solution must be real.
Solution Let r > 1, let f(z) = – λ + z and g(z) = e–z. Then on [–iR, iR],
we have, z = iy & | f(z)| = | – λ + iy | =
λ 2 + y 2 ≥ λ > 1 and | g(z)| = e–z.
= | e–iy| = 1.
Thus | g(z)| < | f(z)| on [–iR, iR].
Now consider the semi circle { z / | z | = R : Re(z) > 0}.
i.e., z = Reit ,
– π/2 < t < π/2.
Here | f(z)| = | – λ + Reit | ≥ R – λ, and | g(z)| = | e–z | | e–R cos t | ≤ 1 ≤ R – λ ≤ | f(z)|.
Thus | g(z)| < | f(z)| on the semi circle & [–iR, iR].
Thus f and f + g have the same number of zeros, inside the circle
{ z / | z | = R ; Re(z) > 0} ∪ [–iR, iR].
Since f(z) = – λ + z has precisely one zero so does f(z) + g(z) = λ – z – e–z.
101
Example 10 Consider z3 – 6z2 + 3z +1,
Solution Let f(z) = 6z2 – 3z
&
g(z) = z3 + 1
Let γ : | z | = 3 and γ = { z / | z | = 3 } .
Now | f(z) | = | 6z2 – 3z | ≥ 6| z |2 – 3| z | = 6×9 – 3×3 = 45
& | g(z) | = | z3 – 1 | = | z |3 + 1 = 28
Thus | g(z) | = | f(z) | for z ∈ { γ }
Hence f and f + g have the same no of zeros in B(0; 3)
Since f(z) = 3z( 2z – 1) both the zeros viz., 0 and ½ of f lies inside | z | = 3
Hence z3 – 6z2 + 3z + 1 = 0 has precisely 2 zeros in B(0 ; 3).
Example 11 Prove that e–iz = 2 + z2 has only one root in the upper half plane.
Solution Let f(z) = 2 + z2 & g(z) = e–iz ,
Consider γ = [–R, R] ∪ {z / Re(z) > 0, | z | = R}
for z ∈ [–R, R], z = x real,
f(z) = f( x) = 2 + x2 > 1,
= | e–ix |
= | e–iz |
= | g(z) |
∴ g(z) < f( z) for z ∈ [–R, R],
∴ let z = Re–it
0 ≤ t ≤ 2π
Then | f(z) | = | 2z + z2 | ≥ R2 – 2
By choosing R > √3 we have
| f(z) | ≥ R2 – 2 > 1 > e–Rsin t = e
iR (cos t + i sin t )
= | e–iz | on {z / Re(z) > 0, | z | = R}
102
Thus | g(z) | ≤ | f( z) |
Hence by Rouches theorem f and f + g have the same no of zeros in γ,
Since f has only one zero viz. √2i inside γ and above real axis i.e. upper half plane f + g has
precisely one zero in the upper half plane.
Maximum Modulus Theorems
Theorem 12 If f is analytic in a region G and a is a point in G with | f(x) | ≥ | g(x) | ∀ z ∈ G,
then f must be a constant function.
Proof : Let α = f(a) & Ω = f (G). By hypothesis
| f(a) | ≥ | f(z) | ∀ z ∈ G ⇒ | α | ≥ | ξ | ∀ z ∈ Ω. Hence Ω ∩ ∂Ω
≠ φ, since α | ≥ | ξ | ∀ ξ
∈ Ω ⇒ α ∈ Ω. Therefore, Ω cannot be an open set. Since f is analytic and non constant Ω
= f (G) is necessarily open by open mapping theorem. Hence f must be a constant.
Maximum Modulus Theorem 13 (2nd version) : Let G be a neighbourhood open set in G
and suppose F is continuous function onG which is analytic in G. Then max
{| f(z) | : z∈G } = {| f(z) | : z ∈∂G}.
Proof : Since G is bounded, G id bounded closed set. Hence,G is compact. Since f is
continuous on compact set G, it attains maximum at some a ∈ G, i.e., there is a∈ G such
that | f(z) | ≤ | f(a) | ∀ z ∈ G. If f were non-constant and a ∈ G, then we are lead to
contradiction by 1st version of MMT, hence a ∈ G but a ∉ G i.e., a ∈∂G.
Definition 14 Let f be real valued function G ⊆ **. Let a ∈ G or a = ∞ then the limit
superior of f(z) as z → ∞ is defined as
lim sup f ( z ) = lim sup f ( z ) / z ∈ G ∩ B (a; r ) .
r →a
r →o +
103
Similarly, lim inf f ( z ) = lim inf f ( z ) / z ∈ G ∩ B (a; r ) .
r →a
r →o +
By ∂∞G, we mean extended boundary of G defined as ∂∞G = ∂G ∪ {∞ }, if g is unbounded &
∂∞G = ∂G if G is bounded.
Maximum Modulus Theorem 15 (3rd version) : Let G be region in C & f an analytic
function on G. Suppose there is constant M s. t. lim sup | f ( z ) |≤ M ∀ a ∈ G. Then | f(z) | ≤ M
r →a
∀ z ∈ G.
Proof: Let δ > 0 be arbitrary and H = {z ∈ G :| f ( z ) |> M + δ } . Now, it is enough to prove
that H = φ . Since f is analytic, its real and imaginary parts are continuous and hence | f | ,
is continuous. Therefore, H is open . Now, given that lim sup | f ( z ) |≤ M
r →a
Hence, there is r > 0 such that | f ( z ) |< M + δ
∀a ∈ ∂ ∞G .
∀z ∈ G ∩ B(a; r ) . Hence, H ∩ ∂ ∞G = φ . In
other words, H ⊆ G , regardless of whether G is bounded or unbounded. Hence, by 2nd
version of MMT, there is z ∈ ∂H , such that | f ( z ) |= M + δ , which is absurd, hence, H = φ
or f is constant. Now if f is constant, H = φ is hypothesis. Thus in any case H = φ .
Schwarz’ Lemma 16 Let D = {z :| z |< 1} and suppose f is analytic in D with
(a) | f ( z ) |≤ 1 for z ∈ D
(b)
f (0) = 0 .
Then | f '(0) |≤ 1, | f ( z ) |≤| z | ∀z ∈ D .
Moreover, if | f '(0) |= 1 or | f ( z ) |=| z | for some z ≠ 0 . Then there is a constant c,| c |= 1 such
that f ( w) = cw, ∀w ∈ D .
104
Proof. Define
lim g ( z ) = lim
z →0
z →0
g:D→
by
g ( z) =
f ( z)
if
z
f ( z)
f ( z ) − f (0)
= lim
= f '(0). Thus
z →0
z
z −0
z≠0
g
and
is
g (0) = f '(0) . Note that
continuous
z=0
at
and
consequently g is analytic in D . Now by maximum modulus theorem, for any r < 1 , there is
a point z0 such that | z0 |= r < 1 and | g ( z ) |=
| g ( z ) |≤ 1 ∀z ∈ D . Hence,
f ( z0 ) 1
f ( z)
≤
≤ . Letting r → 1− , we obtain
z
z0
r
f ( z)
≤ 1 or | f ( z ) |≤| z | for all z ≠ 0 and | g (0) |=| f '(0) |≤ 1 .
z
Since f (0) = 0 , we have | f ( z ) |≤| z | for all z ∈ D and | f '(0) |≤ 1 . Let | f '(0) |= 1 , then
| g (0) |= 1 , which implies that g attains maximum in the disc D. Therefore, g must be
constant. Therefore, there is c ∈
such that g ( z ) = c . That is,
f ( z)
= c or f ( z ) = cz , since
z
f ( z ) = cz holds trivially for z = 0 , we have f ( z ) = cz for all z ∈ D . Further, | c |=| g ( z ) |= 1 .
In other words | c |= 1 ⇒ c = eiθ for some real θ . Hence, f ( z ) = eiθ z for all z ∈ D .
EXERCISES
1.
2.
Find the number of zeros of the following polynomials
i)
z 4 + 6 z 2 + z + 2 in the unit disk and in the ann(0;1,3)
ii)
z 5 − 12 z 2 + 14 in ann(0;1,5 / 2) and also in {z :| z |< 2}
Suppose R > 0 , is sufficiently large, then if p is a polynomial of degree n>0. Then
∫
| z| = R
p '( z )
dz = 2iπ n .
p( z )
105
We, the authors, are not claiming any originality of the content of this book. The
content is taken form the following references.
REFERENCES
1)
J. B. Conway Foundations of One Complex Variable, Springer Int. Student Edition
1973; Indian Reprint by Narosa Publishing House, New Delhi, 1997.
2)
Ponnusamy. Foundations of Complex Analysis, Narosa Publishing House, New
Delhi.
3)
B. Chauhdary. The Elements of Complex Analysis, Newage International.
4)
Shanti Narayan Theory of Functions of A Complex Variables, S. Chand and
Company Ltd. New Delhi, 1986.
5)
L. V. Ahlfors Complex Analysis, Mc Graw Hill Book Co. 1966.
106
UNIT - IX
SCHWARZ’S LEMMA AND ITS CONSEQUENCES
Theorem 9.1 (Schwarz’s Lemma)
Let D = {z : z < 1} and suppose f is analytic on D with
(a)
f ( z ) ≤ 1 for z in D
(b)
f (0) = 0
Then f ' ( 0 ) ≤ 1 and f ( z ) ≤ z for all z in the disk D. Moreover if
f ' ( 0 ) = 1 or if
f ( z ) = z for some z ≠ 0 then there is a constant c, c = 1 , such that f ( w ) = cw for all w in D.
Proof : Let D = {z : z < 1} and f : D → £ is analytic with
(a)
f ( z ) ≤ 1 for z ∈ D
(b)
f (0) = 0
Define g : D → £ by
 f (z)
for z ≠ 0

g (z) =  z
 f ' ( 0) for z = 0

Then g is analytic in D, and hence analytic on B ( 0; r ) = {z : z r} for every 0 < r < 1 .
Applying maximum modulus theorem to g on B (0, r), for all z ∈ B ( 0, r ) we have
g ( z ) ≤ max g ( z ) = max g ( z )
z ≤r
= max
z =r
z =r
f ( z) 1
≤ ,
z
r
by condition (a).
Letting r → 1 we have
g ( z ) ≤ 1 , for all z ∈ B ( 0,1) = D
⇒
f (z)
≤ 1 , ∀z ∈ D , z ≠ 0
z
107
.... (1)
Since f ( 0 ) = 0 , we have
f ( z ) ≤ z , ∀z ∈ D
Further, from (1) we have
f ' (0) = g (0) ≤ 1
Now, if f ' ( 0 ) = 1 or f ( z ) = f ( z ) for some z ≠ 0 in D then
g ( 0) = f ' (0) = 1
or
g (z) =
f (z)
= 1 , for some z ≠ 0 in D.
z
Thus by (1)
g ( w ) ≤ g ( 0 ) , ∀w ∈ D
or
g ( w ) ≤ g ( z ) ; ∀w ∈ D and some z ≠ 0 in D.
Therefore by maximum modulus theorem g must be constant function on D.
This implies g ( w) = c , ∀w ∈ D and some constant c.
⇒
f ( w)
= c , ∀w ∈ D , w ≠ 0 .
w
⇒ f ( w) = cw , ∀w ∈ D , w ≠ 0 .
Since 0 = f ( 0) = c ( 0) , we have
f ( w ) = cw , ∀w ∈ D
where c = g ( 0 ) = f ' ( 0 ) = 1
Theorem 9.2 :
Let D = {z : z < 1} = B ( 0,1) be the unit disk, and ∂D = {z : z < 1} .
Fix a ∈£ such that a < 1 . Define the Mobius transformation.
φa z =
z −a
1 − az
108
Then :
(a)
φ a is a one-one map of D onto itself,
(b)
φ a is analytic in an open disk containing the closure of D,
(c)
the inverse of φ a is φ−a ,
(d)
φ a maps ∂D onto ∂D ,
(e)
φa ( a ) = 0 , φa ( 0 ) = − a
(f)
φa ( a ) = 0 , φa ' ( 0 ) = 1 − a 2 and φa ( a ) =
1
'
1− a
2
.
Proof :
Let D = {z : z < 1} and ∂D = {z : z = 1} .
Fix a ∈£ such that a < 1 .
Consider the Mobius transformation
φa ( z ) =
(a)
z −a
1 − az
We know Mobius transformation is composition of translations, dilations and the inversion.
Since each function involved in the composition of Mobius transformation is bijective on £ ∞ ,
it follows that φ a is bijective on D.
(b)
The function φ a is well defined everywhere except at z =
1
.
a
1
1
= >1.
As a < 1 , we have
a
a
This implies φ a is well defined on an open disk at the origin containing not only D but ∂D also.
Thus φ a is analytic in an open disk containing the closure of D, as D = D U ∂D .
(c)
If a < 1 then −a < 1 . For any z ∈ D
 z−a 
φ− a ( φa ( z ) ) = φ− a 

 1 − az 
 z−a  ( )

 − −a
 1 − az 
=
 z −a 
1 − (− a ) 

 1 − az 
109
 z −a 

+a
 1 − az 
=
 z −a 
1+ a 

 1 − az 
=
z − a + a − aaz
1 − az + az − aa
2
=
z− a z
1− a
2
=
z (1 − a
1− a
2
)
2
=z
Similarly one can show that
φa (φ− a ( z ) ) = z
Thus
φa (φ− a ( z ) ) = z = φ− a (φ a ( z ) )
This proves the inverse of φ a is φ−a . Further φ a maps D onto itself in a one-one fashion.
(d)
Let any z ∈∂D , then z = 1 .
Let z = eiθ , for some real θ .
Then
φa ( z ) = φa ( e iθ )
=
=
eiθ − a
e iθ − a
=
1 − aeiθ
eiθ e −iθ − a
eiθ − a
( eiθ
−a)
=1
[Q w = w for any w∈£ ]
⇒ φa ( z ) ∈ D
This proves φ a maps ∂D on to ∂D .
(e)
From the definition of φ a , it follows that φa ( a ) = 0 , φa ( 0 ) = − a .
(f)
Already we have proved that φ a is analytic in an open disk containing the closure of D.
110
Thus,
φa ' ( z ) =
=
d  z −a 


dz  1 − az 
(1 − az ) (1) − ( z − a ) ( −a )
(1 − az )2
1 − az + az − a
=
(1 − az )2
2
2
=
1− a
(1 − az )2
In particular,
φa ' ( 0 ) = 1 − a
φa ' ( a ) =
2
1− a
2
(1 − aa )
2
=
1− a
2
(1 − a )
2 2
=
1
1− a
2
The following theorem help to find an upper bound for f ' ( z ) of any analytic map f : D → D ,
D = {z : z < 1} .
Theorem 9.3 : Let f is analytic on D with f ( z ) ≤ 1 and let f ( a ) = α for
2
1− α
a ∈ D = { z : z < 1} then f ' ( a ) ≤
2
1− a
Furthermore if
f ' ( a) =
1− α
2
1− a
2
then there is a constanmt c with c = 1 and
f ( z ) = φ −α ( cφa ( z ) ) for z ∈ D .
Proof : Let f is analytic on D with f ( z ) ≤ 1 .
Let a ∈ D = { z : z < 1} and f ( a ) = α , so α < 1 unless f is constant.
Define g = φα o f o φ− a . Then g maps D into D and also satisfies
111
g ( 0 ) = (φα o f o φ− a ) ( 0 )
= φα ( f (φ −a ( 0 ) ) )
= φα ( f ( a ) )
= φα (α )
=0
Thus by applying Schwarz’s Lemma we obtain
g ' (0) ≤ 1
Now by applying chain rule
g ' ( 0 ) = (φα o f o φ− a ) ( 0 )
= ( φα o f ) ' ( φ− a ( 0 ) ) φ− a ' ( 0 )
= ( φα o f ) ' ( a ) (1 − a
2
)
2
= φα ' ( f ( a ) ) f ' ( a ) (1 − a )
= φα ' (α ) f ' ( a ) (1 − a
2
)
f ' ( a ) (1 − a
2
)
=
1
2
1− α
Therefore,
f '( a) =
1− α
2
1− a
2
g ' (0 )
Using the fact g ' ( 0 ) ≤ 1 , we obtain
f ' ( a) ≤
1− α
2
1− a
2
, a ∈D .
Further observe that, equality holds exactly when g ' ( 0 ) = 1 .
Thus applying Schwarz’s Lemma to a function g, there is a constant c, c = 1 , such that.
g ( z ) = cz for all z ∈ D .
112
⇒ (φα o f o φ− a ) ( z ) = cz , ∀z ∈ D
⇒ φα o f o φ− a = cI where I : D → D , I ( z ) = z
⇒ φα o f = cI o φα = cφ∞
⇒ f = φ−α o ( cφa )
⇒ f ( z ) = φ−α ( cφa ( z ) ) for z ∈ D
Theorem 9.4 : Let f : D → D , be a one-one analytic map of D = {z : z < 1} onto itself and
suppose f ( a ) = 0 . Then there is a complex number c with c = 1 such that f = cφα .
Proof : Let f : D → D be a bijective map, where D = {z : z < 1} .
Then there is an analytic function g : D → D such that
g ( f ( z ) ) = z for ∀z ∈ D
.... (1)
Let f ( a ) = 0 . Then by Theorem 9.3
2
1− 0
1
f ' ( a) ≤
=
2
2
1− a
1− a
..... (2)
Further by (1), we have
g ( f ( a)) = a
⇒ g ( 0 ) = a , as f ( a ) = 0
Again applying the Theorem 9.3 to g we obtain
g ' (0) ≤
1− a
2
1− 0
2
= 1− a
2
Differentiating (1), we get
g ' ( f ( z )) f ' ( z) = 1
Therefore,
1 = g ' ( f ( a ) ) f ' (a )
[Q f ( a ) = 0]
= g ' (0) f ' (a )
113
⇒ f ' ( a) =
1
1
≥
g ' ( 0) 1 − a 2
.... (3)
By (2) and (3) we have
f ' ( a) =
1
1− a
2
=
1− 0
2
1− a
2
Therefore by Theorem 9.3, there is a complex number c with c = 1 such that
f ( z ) = φ −0 ( cφa ( z ) )
= φ0 ( cφa ( z ) )
= cφa ( z ) , z ∈ D
⇒ f = cφ a
Problem 9.1 : Let f : D → £ is analytic with Re f ( z ) ≥ 0 for all z in D = {z : z < 1} , and f ( 0 ) = 1.
Show that Re f ( z ) > 0 .
and
1− z
1+ z
≤ f ( z) ≤
,
1+ z
1− z z ∈ D
Solution : Let D = {z : z < 1} , f : D → £ is analytic with Re f ( z ) ≥ 0 for all z ∈ D , and f ( 0 ) = 1.
Define φ ( z ) =
z −1
. Then φ maps { z :Re z > 0} onto D, so the function g = φ o f maps D
z +1
to itself and
g ( 0 ) = φ ( f ( 0 ) ) = φ (1) = 0
Applying Schwarz’s Lemma to g, we obtain
g ( z ) ≤ z , ∀z ∈ D
⇒ φ ( f ( z ) ) ≤ z , ∀z ∈ D
⇒
f
f
( z ) −1
≤ z ; ∀z ∈ D
( z ) +1
114
Therefore
f ( z ) −1 f ( z ) −1
≤
≤ z , ∀z ∈ D
f ( z ) +1 f ( z ) +1
⇒ f ( z ) −1 ≤ z f ( z ) + z
⇒ f ( z ) (1 − z ) ≤ 1 + z
1+ z
⇒ f (z) ≤
;
1 − z ∀z ∈ D
On the same line
⇒
implies
1− f ( z )
f ( z ) −1
≤
≤ z , ∀z ∈ D
1 + f ( z)
f ( z ) +1
1− z
≤ f ( z ) , ∀z ∈ D
1+ z
Therefore
1− z
1+ z
≤ f ( z) ≤
,
1+ z
1 − z ∀z ∈ D
Problem 9.2 : Suppose f ( z ) ≤ 1 for z < 1 and f is analytic. Prove that
f (0) + z
f (z) ≤
1 + f ( 0 ) z for z < 1 ,
Proof : Let f ( z ) ≤ 1 for z < 1 and f is analytic on D = {z : z < 1} .
Let f ( 0 ) = a and consider the function g = φa o f .
Then g ( 0 ) = φa ( f ( 0 ) ) = φa ( a ) = 0
Thus by applying Schwarz’s Lemma, we have
g ( z ) ≤ z , ∀z ∈ D
..... (1)
⇒ φa ( f ( z ) ) ≤ z ; ∀z ∈ D
⇒
f (z) − a
≤ z ; ∀z ∈ D
1 − af ( z )
115
f (z) − a
For any z ∈ D , g ( z ) =
1 − af ( z )
⇒ g ( z ) − ag ( z) f ( z ) = f ( z ) − a
⇒ g ( z ) + a = f ( z ) (1 + ag ( z ) )
⇒ f ( z) =
a +g ( z )
1 + ag ( z )
Thus f ( z ) is obtained from w = g ( z ) by a bilinear transformation, which maps circles onto
circles and centre of a circle onto centre of its image. From (1) we see that for any z ∈ D , g ( z ) is in
the disk B ( 0, z ) . This disk is mapped on a disk D1 ⊆ D with centre f ( a ) = a .
That is, f ( z ) ∈ D1 and f ( z ) ≤ p − 0 , where p is the point on the closure of D1 .
a + z eit
Then P is given by P =
1 + a z e it
Therefore,
a + z eit
f (z) ≤
1 + a z e it
≤
a+z
1+ a z
=
f (0) + z
1 + f ( 0 ) z , for z < 1
[Q a = a ]
EXERCISE :
1.
Suppose f ( z ) ≤ 1 for z < 1 and f is analytic on D = {z : z < 1} .
Prove that :
(a)
f ( 0) + z
f (z) ≤
1 − f ( 0 ) z , for z < 1 .
(b)
1 − f ( 0) z
≤ f ( z ) , for z < 1 .
1+ f ( 0) z
❁❁❁
116
UNIT - X
SPACES OF ANALYTIC FUNCTIONS AND THE RIEMANN
MAPPING THEOREM
The Space of Continuous Functions :
Let G is an open set in £ and ( Ω, d ) is a complete metric space then C ( G, Ω ) denotes the
set of all continuous function from G to Ω .
The set C ( G, Ω ) is non-empty, as it always contains the constant functions.
Theoerm 10.1 : If G is an open set in £ then there is a sequence { K n} of compact subsets of G such
∞
that G = U K n .
n =1
Moreover, the sets can K n be chosen to satisfy the following conditions :
(a)
Kn ⊆ Kn+1 ;
(b)
K ⊂ G and K compact implies K ⊂ Kn for some n;
(c)
Every component of £ ∞ − K n contains a component of £ ∞ − G .
Definition 10.1 :
∞
Let G = U K n , where each K n is compact and Kn ⊂ int K n+1 .
n =1
Define, σ n ( f , g ) = sup {d ( f ( z ) ,g ( z ) ) : z ∈ K n }
for all functions f and g in C ( G, Ω ) .
∞
 1  σ n ( f , g)
σ ( f ,g) = ∑ 
Z
n =1  2  1 + σ n ( f , g )
n
Also define,
....... (1)
∞
t
1
≤ 1 for all t ≥ 0 , the series in (1) is dominated by ∑   and hence it is convergent.
As
1+ t
n =1  2 
n
Theorem 10.2 : ( C ( G , Ω ) , σ ) is a complete metric space.
117
Spaces of Analytic Functions :
Let G be an open subset of the complex plane. Let H ( G ) is the collection of analytic functions
on G. Then we consider H ( G ) as a subset of C ( G, £ ) , and the metric on H ( G ) is the metric
which it inherits as subset of C ( G, £ ) .
Theoerm 10.3 : If
{ fn} is a sequence in H (G) and f
belongs to C ( G, £ ) such that f n → f then
f is analytic and f n(k ) → f (k ) for each integer k ≥ 1 .
Proof : Let { f n} is a sequence in H ( G ) and f ∈ C ( G, £ ) such that f n → f .
To prove f is analytic we use Morera’s theorem. Let T be a triangle contained inside a disk
D⊂G.
Since T is compact, f n → f uniformly over T..
Therefore,
lim
n→∞
∫ fn = ∫ f
T
...... (1)
T
As each f n is analytic on G and T is a closed rectifiable curve in a disk D, by Cauchy’ss
theorem.
∫ fn = 0 , for each n.
T
Therefore from (1) we have,
∫f =0
T
Thus f must be analytic in every disk D ⊂ G , by Morera’s theorem, and hence f is analytic in
G. Now we prove that f n(k ) → f (k ) for each integer k ≥ 1 .
Let D = B (a ; r ) ⊂ G . Choose R > r such that B (a ; R ) ⊂ G .
Let γ is the circle z − a = R , then by Cauchy’s integral formula, for each integer k ≥ 1 , and
each n, we have
fn
(k ) (
z)− f
(k) (
z) =
k ! f n ( w ) − f ( w)
dw ,
z∈D
2π i ∫r ( w − z )k +1
Since {γ } is compact, uniformly on {γ } .
Let M n = sup { f n ( w) − f ( w) : w ∈ {γ }} ,
118
...... (2)
then M n → 0 , as f n → f .
Further for z ∈ D and w ∈ {γ } , we have
w − z = ( w − a ) + (a − z )
≥ w− a − z −a
≥ w− a − z − a
≥ R−r
⇒
1
1
≤
w− z R − r
.... (3)
Thus from (2) and (3), for z ∈ D , we have
f n ( k ) ( z ) − f (k ) ( z ) ≤
k ! f n ( w) − f ( w )
dw
K +1
2π ∫r
w− z
≤
k !M n
dw
k +1 ∫
2π ( R − r ) r
=
k !M n
⋅ 2π R
k +1
2π ( R − r )
=
k !MnR
( R − r ) k +1
Therefore for each Z ∈ D ,
lim f n(k ) ( z ) − f ( k ) ( z ) =
n→∞
k !R
lim M n = 0
( R − r ) k +1 n→∞
Hence f n(k ) → f (k ) uniformly on D = B (a , r ) .
Let k be an arbitrary compact subset of G. Let 0 < r < d ( K , ∂G ) , then there are a1 , a 2 , ....,
a n in K such that
n
K ⊆ U B (a j ; r )
j =1
Since f n(k ) → f (k ) uniformly on each B ( a j ; r ) , the convergence is uniform on K.
As K is an arbitrary compact subset of G, f n(k ) → f (k ) uniformly on G..
119
Corollary 10.1 : H ( G ) is a complete metric space.
Proof : As H ( G ) is collection of analytic functions on G and C ( G, £ ) is the set of all continuous
functions, H ( G ) is subset of C ( G, £ ) .
Thus H ( G ) is a subspace of comlete metric space C ( G, £ ) with metric induced on H ( G )
by metric on C ( G, £ ) .
To prove H ( G ) is complete metric space it suffices to show that H ( G ) is closed in C ( G, £ ) .
Let { f n} be any sequence in H ( G ) converging to f, f ∈ C ( G, £ ) .
By Theorem 10.3 f : G → £ is analytic on G, and hence f ∈ H ( G ) .
This proves H ( G ) is closed in complete metric space C ( G, £ ) , hence H ( G ) is complete.
Corollary 10.2 :
If f n : G → £ is analytic and
f
(k ) (
∞
∑ f n ( z ) converges uniformly on compact sets to
n =1
f ( z ) then
∞
z ) = ∑ f n (k ) ( z )
n =1
Proof : Let f n : G → £ is qualatic
∞
Let
∑ f n ( z ) converges uniformly on compact sets to
n =1
f ( z) .
n
Define Sn ( z ) = ∑ f j ( z ) , then by assumption Sn → f uniformly on compact sets.
j =1
Therefore, by Theorem 10.3
Sn( k ) → f ( k ) uniformly on compact set.
⇒f
(k ) (
n
∞
j =1
j =1
z ) = lim Sn( k ) ( z ) = lim ∑ f j ( z ) = ∑ f j ( z )
n →∞
n →∞
120
Theoerm 10.4 (Hurwitz’s Theorem) :
Let G be a region and suppose the sequence
{ f n}
in H ( G ) converges to f. If f ≡ 0 ,
B (a ; R ) ⊂ G , and f ( z ) ≠ 0 for z − a = R then there is an integer N such that for n ≥ N , f and
f n have the same number of zeros in B (a ; R ) .
Proof : Let G be a region and let the sequence { f n} in H ( G ) converges to f.
Let f ≡ 0 , B (a ; R ) ⊂ G and f ( z ) ≠ 0 for z − a = R .
Define δ = inf { f ( z ) : z − a = R}
As f ( z ) ≠ 0 for z − a > R , we have δ > 0 .
Since f n → f uniformly on {z : z − a = R} , corrosponding to
δ
there is an N ∈ ¥ such
2
that if n ≥ N and z − a = R then
f ( z ) − fn ( z ) <
δ
< δ ≤ f ( z ) ≤ f ( z ) + fn ( z )
2
Therefore, by Rouche’s theorem f and f n have the same number of zero’s in B (a ; R ) .
Corollary 10.3 :
{ f n} ⊂ H ( G )
converges to f in H ( G ) and each f n never vanishes on G then either
f ≡ 0 or f never vanishes.
If
Proof : Let { f n} ⊆ H ( G ) , f ∈ H ( G ) and f n → f .
Let each f n never vanishes on G..
If f is not identically zero, then f ( a ) = 0 for some a ∈ G . Since zeros of an analytic function
are isolated, there is R > 0 such that B (a , R ) ⊆ G such that f ≡ 0 on B (a , R ) .
Therefore by Hurwitz’s Theorem there is an integer N such that for n ≥ N , f and f n have the
same number of zeros in B (a ; R ) .
This is contradiction to the assumption each f n never vanishes on G..
Definition 10.3 (Normal Family) : A set F ⊂ C ( G, £) is normal if each sequence in F has a
subsequence which converges to a function f in C ( G, £ ) .
121
Definition 10.4 (Equicontinuous Family) :
A set F ⊆ C ( G, £) is equicontinuous at a point z0 in G iff for every ε > 0 there is a δ > 0
such that
if z − z0 < δ then d ( f ( z ) , f ( z0 ) ) < ε , for all f ∈ F .
We say that F is equicontinuous on E ⊂ G if for every ε > 0 there is a δ > 0 such that
if z , w ∈ E and z − w < δ then d ( f ( z ) , f ( w ) ) < ε , for all f in F .
Remark 10.1 :
1.
If F consists of single function f then the statement that F is equicontinuous at z0 is only the
statement that f is continuous at z0 .
2.
F = { f } is equicontinuous over E is equivalent to saying that f is uniformly continuous on E.
Definition 10.5 (Locally bounded family) :
A family F ⊂ H ( G ) is locally bounded if for each a ∈ G , there are constants M and r > 0
such that for all f ∈ F ,
f ( z ) ≤ M , for z − a < r .
Alternatively, F is locally bounded if there is an r > 0 such that,
sup { f ( z ) : z − a < r , f ∈ F } < ∞ .
That is, F is locally bounded if about each point a in G there is a disk on which F is
uniformly bounded.
We state few theorems without proof which are required to prove Montel’s theorem and its
consequences.
Theorem 10.5 :
A set F ⊂ C ( G, £) is normal iff its closure is compact.
122
Theorem 10.6 (Arzela-Ascoli Theorem) :
As set F ⊆ C ( G, £) is normal iff the following two conditions are satisfied.
(a)
(b)
for each z in G, { f ( z ) : f ∈ F } has compact closure in Ω ;
F is equicontinuous at each point of G..
Theorem 10.7 : A set F in H (G) is locally bounded iff for each compact set K ⊂ G there is a
constant M such that f ( Z ) ≤ M for all f ∈ F and z in K.
Theorem 10.8 (Montel’s Theorem) : A family F in H (G) is normal iff F is locally bounded.
Proof : Let F ⊆ H ( G ) .
Let F is normal.
We have to prove that F is locally bounded. If possible F is not locally bounded, then there
is a compact set K ⊂ G such that
sup { f ( z ) : z ∈ K , f ∈ F } = ∞
This implies for each n, there is f n ∈F such that f n ( z ) ≥ n for each z ∈ K . That is, there
is a sequence { f n} in F such that
sup { f n ( z ) : z ∈ K } ≥ n
.... (1)
But as F is normal there is a function f ∈ H ( G ) and a subsequence
{ fn } of { fn}
k
such
that f nk → f .
Since K ⊂ G and f ∈ H ( G ) , f is continuous on compact set K, and hence f is bounded on
K. Thus there is M > 0 such that
sup { f ( z ) : z ∈ K} ≤ M
.... (2)
From (1) and (2) we have
{
nk ≤ sup f nk (z ) :z ∈K
}
{
}
{
}
≤ sup f nk ( z) − f ( z ) : z ∈ K + sup { f ( z ) : z ∈ K }
≤ sup f nk ( z) − f ( z ) : z ∈ K + M
123
.... (3)
Since f nk → f , we have
lim sup { f n ( z ) − f ( z ) : z ∈ K } = 0
k →∞
.... (4)
Usince (4) we have
lim nk ≤ M , a contradiction to the fact lim nk = ∞ .
k →∞
k →∞
Therefore F must be locally bounded.
Conversely, let F is locally bounded.
To prove F is normal we use Arzela-Ascoli theorem.
(a)
As F is normal, for each a ∈ G there are constants M and r > 0 such that for all f ∈ F ,
f ( z ) ≤ M , for all z ∈ B (a , r ) .
In particular, f ( a ) ≤ M , ∀f ∈ F .
⇒ A = { f ( a ) : f ∈ F } ⊆ B ( 0; M )
⇒ A ⊆ B( 0; M )
Thus A is closed bounded subset of £ and hence compact by Heine Borel theorem.
We have proved that for each a ∈ G ,
(b)
{ f ( a ) : f ∈F } has compact closure in £ .
We now prove that F is equicontinuous at each point of G..
Fix any a ∈ G .
Let ε > 0 be given.
Since F is locally bounded there are constants M and r > 0 such that B (a ; r ) ⊂ G and
f ( z ) ≤ M for all z ∈ B (a ; r ) , and all f ∈ F .
Let z − a <
r
and f ∈ F then using Cauchy’s formula with γ ( t ) = a + re it , 0 ≤ t ≤ 2π ,
2
we have
1
f ( w)
1
f ( w)
f ( a) − f ( z ) =
dw −
dw
∫
∫
2π i γ w − a
2π i γ w − z
=
1  1
1  ( )
−
f w dw
∫

2π i γ  w − a w − z 
124
=
1  ( w − z ) − ( w− a )  ( )
f w dw
2π i ∫r  ( w − a) ( w − z ) 
=
(a − z)
1
f ( w) dw
∫
2π i r ( w − a ) ( w − z )
Therefore,
1
f (a ) − f ( z ) ≤
2π
≤
=
Choose δ <
{
a−z
f ( w ) dw
w− z
∫ w−a
r
1 M a−z
2π r
2π r ⋅ r
2 2
4M
a−z
r
}
r r
,
ε , then for a − z < δ , we have f ( a ) − f ( z ) < ε for all f ∈ F .
2 4M
Thus F is equicontinuous at each a ∈ G .
We have proved that F satisfies the conditions of Arzela-Ascoli theorem. Therefore F is
normal.
Corollary 10.4 : A set F ⊂ H ( G ) is compact iff it is closed and locally bounded.
Proof : We know the theorems
1)
A set F ⊂ C ( G, £) is normal iff F is compact.
2)
F ⊂ H ( G ) is normal iff F is locally bounded.
Therefore F is locally bounded iff F is compact.
This implies F closed and locally bounded iff F = F is compact.
125
The Riemann Mapping Theorem
Definition 10.6 :
A region G1 is conformally equivalent to the region G2 if there is analytic function f : G1 → £
such that f is one-one and f ( G1 ) = G2 .
This is an equivalence relation.
Definition 10.7 :
An open set G is simply connected if G is connected and every closed rectifiable curve in G is
homotopic to zero.
Theorem 10.9 : Let G be an open connected subset of £ . Then the following are equivalent.
(a)
G is simply connected.
(b)
n ( γ ; q ) = 0 for every closed rectifiable curve γ in G and every point a in G − £ .
(c)
£ ∞ − G is connected.
(d)
For any f ∈ H ( G ) such that f ( z ) ≠ 0 for all z in G , there is a function g ∈ H ( G ) such
that f ( z ) = [ g ( z )] .
2
Theorem 10.10 (Open Mapping Theorem)
Let G be a region and suppose that f is a non-constant analytic function on G. Then for any
open set U in G, f (U) is open.
Theorem 10.11 (Identity Theorem)
Let G be a connected open set and let f : G → £ be an analytic function. Then the following
are equivalent statements.
(a)
f ≡0
(b)
there is a point a in G such that f (n ) ( a ) = 0 for each n ≥ 0 .
(c)
{ z ∈G : f
( z ) = 0} has a limit point in G..
With this basics in our hand we prove the well known Riemann mapping theorem.
126
Theorem 10.12 (Riemann Mapping Theorem)
Let G be a simply connected region which is not the whole plane and let a ∈ G . Then there is
a unique analytic function f : G → £ having the properties :
(a)
(b)
(c)
f ( a ) = 0 and f ' ( a ) > 0 ;
f is one-one.
f ( G ) = {z : z < 1} .
Proof : Let G be a simply connected region which is not the whole plane and let a ∈ G .
Define,
F = { f ∈ H ( G ) : f is one-one, f ( a ) = 0, f ' ( a ) > 0 and f ( G ) ⊂ D}
where D = {z : z < 1}
We give the proof in the following steps.
Step 1 : F is nonempty..
Step 2 : F = F U{0} , that is, if f ∈ F then either f ∈ F or f ≡ 0 on G..
..
Step 3 : There exists f ∈ F such that f ( G ) = D .
Setp 4 : f is unique satisfying the conditions in (a), (b) and (c).
Let us proceed towards the first step.
Step 1 : We prove F is non-empty. As G ≠ £ , G is proper subset of £ . Thus there is b ∈£ such
that b ∉ G .
Since G is simply connected, z − b ∈ H ( G ) and z − b ≠ 0 for all z in G, there is a function
g ∈ H ( G ) such that
[ g ( z ) ]2 = z − b
If z1 and z2 be any point in G then
g ( z1 ) = ± g ( z 2 ) ⇒  g ( z1 )  =  g ( z 2 ) 
2
⇒ z1− b = z2 − b
⇒ z1 = z 2
127
2
In particular, g is one-one and
g ( z1 ) = − g ( z2 ) ⇒ z1 = z 2 , ∀z1, z2 ∈ G
.... (1)
As g is non constant analytic function on G, by open mapping theorem, g ( G ) is open in £ .
Further, a ∈ G implies g ( a ) ∈ g ( G ) .
Therefore there exists r > 0 such that
B ( g (a ); r ) ⊂ g (G )
..... (2)
We claim that g ( z ) + g ( a ) ≥ r , ∀z ∈ G , that is,
B ( − g ( a) ; r ) I g ( G ) = φ
If B ( − g ( a) ; r ) I g ( G ) ≠ φ , there is z ∈ G such that g ( z ) ∈ B ( − g ( a) ; r ) , that is,
g ( z ) + g (a ) < r
⇒ − g ( z ) − g (a ) < r
⇒ − g ( z ) ∈ B ( g ( a); r )
⇒ − g ( z) ∈ g (G )
[Q By (2)]
Therefore, ∃w ∈ G such that − g ( z ) = g ( w) .
But (1) implies that z = w.
Thus g ( z ) = g ( w) = − g ( z ) .
⇒ 2g ( z) = 0
⇒ [ g ( z )] = 0
2
⇒ z−b= 0
⇒z =b
But z = w gives that b = w ∈G , a contradiction.
Thus we must have
B ( − g ( a) ; r ) I g ( G ) = φ
⇒ g ( z ) + g ( a ) ≥ r , ∀z ∈ G
.... (3)
128
Inparticular for z = a ∈ G , we have
g (a ) + g (a) ≥ r ⇒ 2 g ( a) ≥ r
r
2
.... (4)
r g ' (a) g (a) g ( z ) − g ( a) ,
z ∈G
⋅
⋅
⋅
4 g ' (a) g (a ) 2 g ( z ) + g ( a)
..... (3)
⇒ g (a ) ≥
Define, h : G → £ by
h ( z) =
As g ∈ H ( G ) , we have h ∈ H ( G ) .
Further h ( a ) = 0 .
Let z1 , z 2 ∈ G and h ( z1 ) = h ( z 2 )
g ( z1 ) − g ( a ) g ( z2 ) − g ( a )
then g ( z ) + g ( a ) = g ( z ) + g ( a )
1
2
⇒ g ( z1 ) = g ( z 2 )
⇒ z1 = z 2 , as g is one-one.
This proves h is one-one.
From (3), we have
h' z =
r g ' (a ) g ( a) ( g ( z ) + g ( a)) g ' ( z ) − ( g ( z ) − g (a )) g ' (z )
⋅
⋅
⋅
4 g ' (a ) g (a ) 2
[ g ( z ) + g ( a )]2
=
r g ' (a ) g (a )
2g '( z ) g ( z )
⋅
⋅
⋅
2
4 g ' ( a ) g ( a ) [ g ( z ) + g ( a ) ]2
Therefore,
r g ' ( a ) g ( a ) 2 g ' (a )g (a )
⋅
⋅
⋅
4 g ' ( a ) g ( a ) 2 4 [ g ( a ) ]2
h '( a) =
=
r g ' (a )
⋅
>0
8 g ( a) 2
Thus h ' ( a ) > 0 .
129
Now
r g ' (a) g ( a) g ( z) − g ( a)
⋅
⋅
⋅
4 g ' (a) g (a) 2 g ( z ) + g ( a)
h(z) =
=
r
1 g ( z ) − g ( a)
⋅
4 g (a ) g ( z ) + g ( a )
=
r
g ( z ) − g ( a)
4 g (a )[ g ( z) + g ( a)]
=
r [ g ( z ) + g ( a )] − 2 g ( a )
4 g (a )[ g ( z ) + g (a )]
=
r 1
2
−
4 g (a ) g ( z ) + g (a )
≤
r 1
2

+

4  g ( a ) g ( z ) + g ( a ) 
≤
r 2 2
 + 
4r r 
[ Q By (3) and (4)]
= 1
Thus h ( z ) ≤ 1 , ∀z ∈ G .
But h being non consant analytic function G, it cannot attains its maximum on G.
Hence h ( z ) < 1 , ∀z ∈ G .
⇒ h (G ) ⊂ D
We have proved that h ∈ H ( G ) , h is one-one, h ( a ) = 0 , h ' ( a ) > 0 and h ( G ) ⊂ D .
Therefore h ∈F , and consequently, F is nonempty..
Step 2 : In this step we prove that F = F U{0} .
Let any f ∈ F .
Then there is a sequence { f n} in F such that f n → f on G..
130
Therefore,
f ( a ) = lim f n ( a ) and f ' ( a ) = lim f n ' ( a )
h →∞
h→∞
But for each n, f n ∈F implies f n ( a ) = 0 and f n ' ( a ) > 0 .
Thus f ( a ) = 0 and f ' ( a ) ≥ 0
.... (5)
We claim that f ' ( a ) ≠ 0 .
Fix any z1 ∈ G and let z 2 ∈ G such that z1 ≠ z 2 .
Then ∃ε > 0 such that z1 ∉ B ( z2 ; ε ) = K .
Let f ( z1 ) = ξ and f n ( z1 ) = ξn , for each n.
Since f n is one-one for each n, f n − ξ n is never vanishing on K.
As f n → f , f n − ξ n → f − ξ on K.
As K is compact f n − ξ n → f − ξ uniformly on K, so Hurwitz’s theorem gives that f ( z ) − ξ
never vanishes on K or f ( z ) − ξ ≡ 0 .
If f ( z ) − ξ ≡ 0 on K then f ( z ) = ξ on G, that is, f is the constant function ξ throughout G..
Since f ( a ) = 0 ⇒ ξ = 0 , and hence we have f ≡ 0 on G..
On the other hand if f ( z ) − ξ is never vanishing on K, then
z1 ≠ z2 ⇒ f ( z1 ) − ξ ≠ f ( z2 ) − ξ
⇒ f ( z1 ) ≠ f ( z 2 )
Thus f is one-one
But if f is one-one than f ’ can never vanish, so (5) implies f ' ( a ) > 0 .
Next, for each n, f n ∈F implies f n ( G ) ⊂ D , that is f n ( z ) < 1 .
Therefore,
f n ( z ) = lim f n ( z ) = lim f n ( z ) ≤ 1
n →∞
n →∞
If f ( z ) = 1 for some z ∈ G , maximum modulus theorem forces f to be constant.
131
Thus we must have
f ( z ) < 1 , ∀z ∈ G
⇒ f (G ) ⊂ D
Therefore f is analytic on G, f ( a ) = 0 , f ' ( a ) > 0 , f is one-one and f ( G ) ⊂ D .
This proves f ∈ F .
We have proved that for any f ∈ F , either f = 0 or f ∈ F .
Hence F = F U{0} .
Step 3 : We prove that there exists f ∈ F such f ( G ) = D .
Consider the function φ : H ( G ) → £ , defined by
φ ( f ) = f ' ( a)
Let { f n} be any sequence in H ( G ) and f ∈ H ( G ) such that f n → f then f is analytic and
f n(k ) → f (k ) for each integer k ≥ 1 .
Inparticular f n ' → f '⇒ f n ' ( a ) → f ' ( a ) .
⇒ φ ( fn ) → φ ( f )
This proves φ is continuous.
Further, f ( G ) ⊂ D , ∀f ∈ F implies that
sup { f ( z ) : z ∈ D , ∀f ∈ F } ≤ 1
Hence, F is locally bounded an consequently, by Montel’s theorem F is compact.
As φ is continuous on compact set F , there is f ∈ F such that
{
φ ( f ) = max φ ( g ) : g ∈ F
}
⇒ f ' ( a ) = max { g ' ( a ) : g ∈ F }
⇒ f ' ( a ) = max {g '( a) : g ∈ F }
⇒ f ' ( a ) ≥ g ' ( a ) , ∀g ∈ F
.... (6)
As F ≠ φ and F = F U{0} implies f ∈ F .
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For this f we prove that f ( G ) = D .
If f ( G ) ≠ D , then there is w ∈ D such that w ∉ f ( G ) .
Thus w ≠ f ( z ) , ∀z ∈ G .
This implies
f (z) −w
f (z)− w
∈ H ( G ) and
≠ 0 for z ∈ G .
1 − wf ( z )
1 − wf ( z )
Therefore, there is an analytic function h : G → £ such that
( )
[ h ( z ) ]2 = f z − w
..... (7)
1 − wf ( z )
Since the Mobius transformation T ξ =
ξ−w
maps D on D, we have h ( G ) ⊂ D .
1 − wξ
Define g : G → £ by
h ' ( a) h ( z ) − h (a )
⋅
h ' ( a ) 1 − h ( a ) h( z )
g (z) =
then g is analytic and g ( G ) ⊂ D .
Further g ( a ) = 0 , g is one-one and
g '( z) =
h ' ( a ) (1 − h (a )h ( z ) ) h ' ( z ) − ( h ( z ) − h ( a ) ) h (a )h ' ( z )
⋅
2
h '( a)
1 − h ( a ) h( z ) 
=
h ' ( a ) h ' ( z ) (1 − h (a )h ( z ) −h ( z ) h ( a ) + h ( a ) h ( a ) )
⋅
2
h ' (a )
1 − h ( a ) h( z) 
Therefore,
2
h ' ( a ) h ' ( a ) 1 − h ( a ) 
g '( a) =
⋅
2
h '( a)
1 − h ( a ) 2 
h ' (a )
.... (8)
2
1− h ( a)
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But,
h (a ) =
2
f ( a) − w
= −w = w
1 − wf ( a )
.... (9)
Differentiating (7) we obtain
2h( z ) h '( z ) =
=
[1 − wf ( z )] ( f ' ( z ) ) − [ f ( z ) − w] ( −wf ' ( z) )
[1 − wf ( z ) ]2
2
f ' ( z ) 1 − wf ( z ) + wf ( z ) − w 
[1 − wf ( z )] 2
2
f ' ( a ) 1 − w 
2
(
)
(
)
⇒ 2 h a h' a =
= f ' ( a ) 1 − w  , as f ( a ) = 0
2
[1 − wf ( z ) ]
⇒ h '( a) =
f ' ( a ) (1 − w
2h ( a )
2
)
.... (10)
Using (9), (10) in (8) we obtain,
g '( a) =
=
2
f ' ( a ) (1 − w )
2 w
f '( a)
2 w
⋅
⋅
1
(1 − w )
( 1− w )( 1+ w )
1− w
 1+ w 
g '( a) = f '( a) 
 > f ' ( a) > 0
2
w


Thus g ' ( a ) > 0 and g ' ( a ) > f ' ( a ) .
This gives that g ∈ F and contradicts to the choice of f, given in (6).
Hence, we must have f ( G ) = D .
Step 4 : We prove that f is unique satisfying conditions (a), (b) and (c).
Suppose there is analytic function g : G → £ satisfying the conditions (a), (b) and (c).
Then fog −1 : D → D is analytic, one-one and onto.
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Also ( fog −1 ) ( 0 ) = f ( g −1 ( 0 ) ) = f ( a ) = 0 .
Hence, by Schwartz’s lemma, there is a constant c with c = 1 and fog −1 ( z ) = cz for all
z∈D.
As g ( G ) = D , z ∈ G ⇒ g ( z ) ∈ D .
Thus
( fog −1 ) ( g ( z ) ) = cg ( z ) , ∀z ∈ G
⇒ f ( g −1 ( g ( z ) ) ) = cg ( z ) , ∀z ∈ G
⇒ f ( z ) = cg ( z ) , ∀z ∈ G
..... (11)
⇒ 0 < f ' ( a ) = cg ' ( a ) .
But g ' ( a ) > 0 implies c > 0.
Thus 1 = c = c .
Therefore from (11) we have,
f ( z ) = g ( z ) , ∀z ∈ G .
Hence f = g on G. This proves the uniqueness.
❁❁❁
135
Writting Team
Unit Number
Dr. Kishor D. Kucche
Department of Mathematics,
9, 10
Shivaji University, Kolhapur - 416 004
Maharashtra
136